Conversion of expressions containing degrees. Converting expressions
Let us consider the topic of transforming expressions with powers, but first we will dwell on a number of transformations that can be carried out with any expressions, including exponential ones. We will learn how to open parentheses, bring such terms, work with the radix and exponent, and use the properties of degrees.
What are exponential expressions?
IN school course few people use the phrase "exponential expressions", but this term is constantly found in collections to prepare for the exam. In most cases, a phrase denotes expressions that contain degrees in their records. We will reflect this in our definition.
Definition 1
Exponential expression Is an expression that contains degrees.
Here are some examples of exponential expressions, starting with a degree with a natural exponent and ending with a degree with a real exponent.
The simplest power expressions can be considered powers of a number with a natural exponent: 3 2, 7 5 + 1, (2 + 1) 5, (- 0, 1) 4, 2 2 3 3, 3 a 2 - a + a 2, x 3 - 1, (a 2) 3. And also degrees with zero exponent: 5 0, (a + 1) 0, 3 + 5 2 - 3, 2 0. And degrees with negative integer powers: (0, 5) 2 + (0, 5) - 2 2.
It is a little more difficult to work with a degree that has rational and irrational indicators: 264 1 4 - 3 3 3 1 2, 2 3, 5 2 - 2 2 - 1, 5, 1 a 1 4 a 1 2 - 2 a - 1 6 b 1 2, x π x 1 - π, 2 3 3 + 5.
The indicator can be the variable 3 x - 54 - 7 3 x - 58 or the logarithm x 2 l g x - 5 x l g x.
With the question of what power expressions are, we figured out. Now let's get down to converting them.
Basic types of transformations of power expressions
First of all, we will look at the basic identity transformations of expressions that can be performed with exponential expressions.
Example 1
Calculate the value of the exponential expression 2 3 (4 2 - 12).
Solution
We will carry out all transformations in compliance with the order of actions. In this case, we will start by performing the actions in brackets: replace the degree with a numerical value and calculate the difference between the two numbers. We have 2 3 (4 2 - 12) = 2 3 (16 - 12) = 2 3 4.
It remains for us to replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32... Here is our answer.
Answer: 2 3 (4 2 - 12) = 32.
Example 2
Simplify the expression with powers 3 a 4 b - 7 - 1 + 2 a 4 b - 7.
Solution
The expression given to us in the problem statement contains similar terms, which we can give: 3 a 4 b - 7 - 1 + 2 a 4 b - 7 = 5 a 4 b - 7 - 1.
Answer: 3 a 4 b - 7 - 1 + 2 a 4 b - 7 = 5 a 4 b - 7 - 1.
Example 3
Present an expression with powers of 9 - b 3 · π - 1 2 as a product.
Solution
Let's represent the number 9 as a power 3 2 and apply the abbreviated multiplication formula:
9 - b 3 π - 1 2 = 3 2 - b 3 π - 1 2 = = 3 - b 3 π - 1 3 + b 3 π - 1
Answer: 9 - b 3 π - 1 2 = 3 - b 3 π - 1 3 + b 3 π - 1.
And now let's move on to the analysis of identical transformations that can be applied precisely in relation to power expressions.
Working with base and exponent
A degree in the base or exponent can have numbers, variables, and some expressions. For example, (2 + 0, 3 7) 5 - 3, 7 and ... It is difficult to work with such records. It is much easier to replace an expression in the base of a power or an expression in an exponent with the same equal expression.
Conversions of the degree and exponent are carried out according to the rules known to us separately from each other. The most important thing is that as a result of the transformations, an expression identical to the original one is obtained.
The purpose of transformations is to simplify the original expression or obtain a solution to a problem. For example, in the example that we gave above, (2 + 0, 3 7) 5 - 3, 7, you can follow the steps to go to the degree 4 , 1 1 , 3 ... Expanding the brackets, we can give similar terms in the base of the degree (a (a + 1) - a 2) 2 (x + 1) and get an exponential expression of a simpler form a 2 (x + 1).
Using degree properties
Power properties, written as equalities, are one of the main tools for transforming power expressions. Here are the main ones, taking into account that a and b Are any positive numbers, and r and s- arbitrary real numbers:
Definition 2
- a r a s = a r + s;
- a r: a s = a r - s;
- (a b) r = a r b r;
- (a: b) r = a r: b r;
- (a r) s = a r s.
In cases where we are dealing with natural, whole, positive exponents, the restrictions on the numbers a and b can be much less strict. So, for example, if we consider the equality a m a n = a m + n, where m and n Are natural numbers, then it will be true for any values of a, both positive and negative, as well as for a = 0.
It is possible to apply the properties of degrees without restrictions in cases where the bases of the degrees are positive or contain variables, the range of admissible values of which is such that the bases on it take only positive values. In fact, within school curriculum in mathematics, the student's task is to select a suitable property and apply it correctly.
When preparing for admission to universities, there may be problems in which the inaccurate use of properties will lead to a narrowing of the ODZ and other difficulties with the solution. In this section, we will analyze only two such cases. More information on the subject can be found in the topic "Transforming Expressions Using Power Properties".
Example 4
Imagine the expression a 2.5 (a 2) - 3: a - 5.5 as a degree with a radix a.
Solution
First, we use the exponentiation property and transform the second factor by it (a 2) - 3... Then we use the properties of multiplying and dividing powers with the same base:
a 2, 5 a - 6: a - 5, 5 = a 2, 5 - 6: a - 5, 5 = a - 3, 5: a - 5, 5 = a - 3, 5 - (- 5, 5) = a 2.
Answer: a 2.5 (a 2) - 3: a - 5.5 = a 2.
Transformation of exponential expressions according to the property of degrees can be performed both from left to right and in the opposite direction.
Example 5
Find the value of the exponential expression 3 1 3 · 7 1 3 · 21 2 3.
Solution
If we apply the equality (a b) r = a r b r, from right to left, then we get a product of the form 3 · 7 1 3 · 21 2 3 and further 21 1 3 · 21 2 3. Let us add up the exponents when multiplying degrees with the same bases: 21 1 3 21 2 3 = 21 1 3 + 2 3 = 21 1 = 21.
There is one more way to carry out transformations:
3 1 3 7 1 3 21 2 3 = 3 1 3 7 1 3 (3 7) 2 3 = 3 1 3 7 1 3 3 2 3 7 2 3 = = 3 1 3 3 2 3 7 1 3 7 2 3 = 3 1 3 + 2 3 7 1 3 + 2 3 = 3 1 7 1 = 21
Answer: 3 1 3 7 1 3 21 2 3 = 3 1 7 1 = 21
Example 6
The exponential expression is given a 1, 5 - a 0, 5 - 6, enter a new variable t = a 0.5.
Solution
Imagine the degree a 1, 5 how a 0.5 3... We use the property of degree to degree (a r) s = a r s from right to left and we get (a 0, 5) 3: a 1, 5 - a 0, 5 - 6 = (a 0, 5) 3 - a 0, 5 - 6. You can easily enter a new variable into the resulting expression. t = a 0.5: we get t 3 - t - 6.
Answer: t 3 - t - 6.
Converting fractions containing powers
We usually deal with two variants of exponential expressions with fractions: the expression is a fraction with a power or contains such a fraction. All basic transformations of fractions are applicable to such expressions without restrictions. They can be reduced, reduced to a new denominator, and worked separately with the numerator and denominator. Let us illustrate this with examples.
Example 7
Simplify the exponential expression 3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2.
Solution
We are dealing with a fraction, so we will carry out transformations in both the numerator and the denominator:
3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = 3 5 2 3 5 1 3 - 3 5 2 3 5 - 2 3 - 2 - x 2 = = 3 5 2 3 + 1 3 - 3 5 2 3 + - 2 3 - 2 - x 2 = 3 5 1 - 3 5 0 - 2 - x 2
Place a minus in front of the fraction to change the sign of the denominator: 12 - 2 - x 2 = - 12 2 + x 2
Answer: 3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = - 12 2 + x 2
Fractions containing powers are reduced to a new denominator in the same way as rational fractions. To do this, you need to find an additional factor and multiply the numerator and denominator of the fraction by it. It is necessary to select an additional factor in such a way that it does not vanish for any values of the variables from the ODZ variables for the original expression.
Example 8
Reduce fractions to the new denominator: a) a + 1 a 0, 7 to the denominator a, b) 1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 to the denominator x + 8 y 1 2.
Solution
a) Let's choose a factor that will allow us to make a reduction to a new denominator. a 0.7 a 0, 3 = a 0.7 + 0, 3 = a, therefore, as an additional factor, we take a 0, 3... The range of valid values of the variable a includes the set of all positive real numbers. In this area, the degree a 0, 3 does not vanish.
Let's multiply the numerator and denominator of the fraction by a 0, 3:
a + 1 a 0, 7 = a + 1 a 0, 3 a 0, 7 a 0, 3 = a + 1 a 0, 3 a
b) Pay attention to the denominator:
x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 2 - x 1 3 2 y 1 6 + 2 y 1 6 2
Multiply this expression by x 1 3 + 2 y 1 6, we get the sum of cubes x 1 3 and 2 y 1 6, i.e. x + 8 y 1 2. This is our new denominator, to which we need to reduce the original fraction.
So we found an additional factor x 1 3 + 2 · y 1 6. On the range of admissible values of variables x and y the expression x 1 3 + 2 y 1 6 does not vanish, so we can multiply the numerator and denominator of the fraction by it:
1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 + 2 y 1 6 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 3 + 2 y 1 6 3 = x 1 3 + 2 y 1 6 x + 8 y 1 2
Answer: a) a + 1 a 0, 7 = a + 1 a 0, 3 a, b) 1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = x 1 3 + 2 y 1 6 x + 8 y 1 2.
Example 9
Reduce the fraction: a) 30 x 3 (x 0.5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0.5 + 1 2 x + 2 x 1 1 3 - 5 3, b) a 1 4 - b 1 4 a 1 2 - b 1 2.
Solution
a) We use the largest common denominator (GCD), by which the numerator and denominator can be reduced. For numbers 30 and 45, this is 15. We can also reduce by x 0.5 + 1 and on x + 2 x 1 1 3 - 5 3.
We get:
30 x 3 (x 0.5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0.5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 x 3 3 (x 0.5 + 1)
b) Here, the presence of the same factors is not obvious. You will have to perform some transformations in order to get the same factors in the numerator and denominator. To do this, we expand the denominator using the formula for the difference of squares:
a 1 4 - b 1 4 a 1 2 - b 1 2 = a 1 4 - b 1 4 a 1 4 2 - b 1 2 2 = = a 1 4 - b 1 4 a 1 4 + b 1 4 a 1 4 - b 1 4 = 1 a 1 4 + b 1 4
Answer: a) 30 x 3 (x 0.5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0.5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 X 3 3 (x 0, 5 + 1), b) a 1 4 - b 1 4 a 1 2 - b 1 2 = 1 a 1 4 + b 1 4.
The main actions with fractions include converting to a new denominator and reducing fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, first the fractions are brought to a common denominator, after which actions (addition or subtraction) are performed with the numerators. The denominator remains the same. The result of our actions is a new fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators.
Example 10
Follow steps x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2.
Solution
Let's start by subtracting the fractions that are in parentheses. Let's bring them to a common denominator:
x 1 2 - 1 x 1 2 + 1
Subtract the numerators:
x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 + 1 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 - x 1 2 - 1 x 1 2 - 1 x 1 2 + 1 x 1 2 - 1 1 x 1 2 = = x 1 2 + 1 2 - x 1 2 - 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 2 + 2 x 1 2 + 1 - x 1 2 2 - 2 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2
Now we multiply the fractions:
4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 x 1 2
Reduce by the degree x 1 2, we get 4 x 1 2 - 1 x 1 2 + 1.
Additionally, you can simplify the exponential expression in the denominator using the difference of squares: squares formula: 4 x 1 2 - 1 x 1 2 + 1 = 4 x 1 2 2 - 1 2 = 4 x - 1.
Answer: x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = 4 x - 1
Example 11
Simplify the exponential expression x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3.
Solution
We can reduce the fraction to (x 2, 7 + 1) 2... We get the fraction x 3 4 x - 5 8 x 2, 7 + 1.
Let's continue the transformation of the degrees of x x 3 4 x - 5 8 · 1 x 2, 7 + 1. Now you can use the property of division of powers with the same bases: x 3 4 x - 5 8 1 x 2, 7 + 1 = x 3 4 - - 5 8 1 x 2, 7 + 1 = x 1 1 8 1 x 2 , 7 + 1.
We pass from the last product to the fraction x 1 3 8 x 2, 7 + 1.
Answer: x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3 = x 1 3 8 x 2, 7 + 1.
In most cases, it is more convenient to transfer multipliers with negative exponents from the numerator to the denominator and vice versa, changing the sign of the exponent. This action allows you to simplify the further solution. Here's an example: the exponential expression (x + 1) - 0, 2 3 x - 1 can be replaced by x 3 (x + 1) 0, 2.
Converting expressions with roots and powers
In problems, there are power expressions that contain not only powers with fractional indicators but also the roots. It is desirable to reduce such expressions only to roots or only to degrees. Going to degrees is preferable as they are easier to work with. Such a transition is especially preferable when the LDV of variables for the original expression allows replacing the roots with powers without the need to refer to the module or split the LDV into several intervals.
Example 12
Imagine the expression x 1 9 x x 3 6 as a power.
Solution
Variable range x is defined by two inequalities x ≥ 0 and x x 3 ≥ 0, which define the set [ 0 , + ∞) .
On this set, we have the right to go from roots to powers:
x 1 9 x x 3 6 = x 1 9 x x x 1 3 1 6
Using the properties of the degrees, we simplify the resulting exponential expression.
x 1 9 x x 1 3 1 6 = x 1 9 x 1 6 x 1 3 1 6 = x 1 9 x 1 6 x 1 1 3 6 = = x 1 9 x 1 6 X 1 18 = x 1 9 + 1 6 + 1 18 = x 1 3
Answer: x 1 9 x x 3 6 = x 1 3.
Converting exponents with variable in exponent
These transformations are quite simple to carry out if the properties of the degree are used correctly. For example, 5 2 x + 1 - 3 5 x 7 x - 14 7 2 x - 1 = 0.
We can replace the product of the degree, in terms of which there is the sum of a variable and a number. On the left side, this can be done with the first and last terms on the left side of the expression:
5 2 x 5 1 - 3 5 x 7 x - 14 7 2 x 7 - 1 = 0.5 5 2 x - 3 5 x 7 x - 2 7 2 x = 0.
Now we divide both sides of the equality by 7 2 x... This expression on the ODZ of the variable x takes only positive values:
5 5 - 3 5 x 7 x - 2 7 2 x 7 2 x = 0 7 2 x, 5 5 2 x 7 2 x - 3 5 x 7 x 7 2 x - 2 7 2 x 7 2 x = 0.5 5 2 x 7 2 x - 3 5 x 7 x 7 x 7 x - 2 7 2 x 7 2 x = 0
Reducing the fractions with powers, we get: 5 5 2 x 7 2 x - 3 5 x 7 x - 2 = 0.
Finally, the ratio of the degrees with the same indicators is replaced by the powers of the ratios, which leads to the equation 5 · 5 7 2 · x - 3 · 5 7 x - 2 = 0, which is equivalent to 5 · 5 7 x 2 - 3 · 5 7 x - 2 = 0.
Introduce a new variable t = 5 7 x, which reduces the solution to the original exponential equation to the solution of the quadratic equation 5 · t 2 - 3 · t - 2 = 0.
Convert expressions with powers and logarithms
Expressions that contain degrees and logarithms are also found in problems. Examples of such expressions are: 1 4 1 - 5 · log 2 3 or log 3 27 9 + 5 (1 - log 3 5) · log 5 3. The transformation of such expressions is carried out using the approaches and properties of logarithms discussed above, which we discussed in detail in the topic "Converting logarithmic expressions".
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Topic: " Converting expressions containing exponents with fractional exponents "
“Let someone try to erase degrees from mathematics, and he will see that without them you cannot go far.” (M.V. Lomonosov)
Lesson objectives:
educational: to generalize and systematize the knowledge of students on the topic “Degree with rational indicator"; Control the level of assimilation of the material; eliminate gaps in the knowledge and skills of students;
developing: form students' self-control skills; create an atmosphere of interest of each student in work, develop cognitive activity students;
educational: foster interest in the subject, in the history of mathematics.
Lesson type: lesson in generalization and systematization of knowledge
Equipment: grade sheets, cards with assignments, decoders, crosswords for each student.
Preliminary preparation: the class is divided into groups, in each group the leader is a consultant.
DURING THE CLASSES
I. Organizing time.
Teacher: We have completed the study of the topic "Degree with a rational exponent and its properties." Your task in this lesson is to show how you learned the studied material and how you are able to apply the knowledge gained to solving specific problems. Each of you has a score sheet on the table. In it you will enter your grade for each stage of the lesson. At the end of the lesson, you will expose average score per lesson.
Evaluation paper
| Crossword | Warm up | Work in | Equations | Check yourself (s \ r) | ||
II. Examination homework.
Mutual examination with a pencil in hand, the answers are read out by the students.
III. Updating students' knowledge.
Teacher: The famous French writer Anatole France said at one time: "Learning must be fun. ... To absorb knowledge, one must absorb it with appetite."
Let's repeat the necessary theoretical information in the course of solving the crossword puzzle.
Horizontally:
1. The action by which the degree value is calculated (erection).
2. A product consisting of the same factors (degree).
3. The effect of exponents when raising a degree to a degree (work).
4. Action of degrees at which exponents are subtracted (division).
Vertically:
5. The number of all the same factors (index).
6. Degree with zero exponent (unit).
7. Duplicate multiplier (base).
8. Value 10 5: (2 3 5 5) (four).
9. An exponent that is not usually written (unit).
IV. Mathematical warm-up.
Teacher. Let's repeat the definition of the degree with a rational exponent and its properties, we will perform the following tasks.
1. Represent the expression x 22 as a product of two degrees with base x, if one of the factors is: x 2, x 5.5, x 1 \ 3, x 17.5, x 0
2. Simplify:
b) y 5 \ 8 y 1 \ 4: y 1 \ 8 = y
c) s 1.4 s -0.3 s 2.9
3. Calculate and form a word using a decoder.
After completing this task, you guys will learn the name of the German mathematician who introduced the term “exponent”.
1) (-8) 1\3 2) 81 1\2 3) (3\5) -1 4) (5\7) 0 5) 27 -1\3 6) (2\3) -2 7) 16 1\2 * 125 1\3
Word: 1234567 (Pin)
V. Written work in notebooks (answers open on the board) .
Tasks:
1. Simplify the expression:
(x-2): (x 1 \ 2 -2 1 \ 2) (y-3): (y 1 \ 2 - 3 1 \ 2) (x-1): (x 2 \ 3 -x 1 \ 3 +1)
2. Find the value of the expression:
(x 3 \ 8 x 1 \ 4 :) 4 at x = 81
Vi. Group work.
The task. Solve equations and form a word using a decoder.
Card number 1
Word: 1234567 (Diophantus)
Card number 2
Card number 3
Word: 123451 (Newton)
Decoder
Teacher. All these scholars have contributed to the development of the concept of "degree".
Vii. Historical information about the development of the concept of degree (student message).
The concept of a degree with a natural indicator was formed even among the ancient peoples. Square and cube numbers were used to calculate areas and volumes. The degrees of some numbers were used by scientists to solve certain problems. Ancient egypt and Babylon.
In the III century, the book of the Greek scientist Diophantus "Arithmetic" was published, which laid the foundation for the introduction of alphabetic symbolism. Diophantus introduces symbols for the first six powers of the unknown and their reciprocal values. In this book, a square is denoted by a sign with an index r; the cube is by the sign k with the index r, and so on.
From the practice of solving more complex algebraic problems and operating with degrees, it became necessary to generalize the concept of a degree and expand it by introducing zero, negative and fractional numbers as an exponent. The idea of generalizing the concept of a degree to a degree with an unnatural exponent of mathematics came gradually.
Fractional exponents and the simplest rules of action on powers with fractional exponents are found in the French mathematician Nicholas Orem (1323–1382) in his work “Algorithm of Proportions”.
Equality, and 0 = 1 (for and not equal to 0) was used in his writings at the beginning of the 15th century by the Samarkand scientist Giyasaddin Kashi Dzhemshid. Independently of him, the zero indicator was introduced by Nikolai Shuke in the 15th century. It is known that Nikolai Shuke (1445–1500) considered degrees with negative and zero exponents.
Later, fractional and negative exponents are found in “Complete arithmetic” (1544) by the German mathematician M. Stiefel and in Simon Stevin. Simon Stevin suggested to mean a 1 / n root.
The German mathematician M. Stiefel (1487–1567) defined a 0 = 1 at and introduced the name of the exponent (this is a literal translation from the German Exponent). German potenzieren means exponentiation.
At the end of the sixteenth century, François Viet introduced letters to denote not only variables, but also their coefficients. He used abbreviations: N, Q, C - for the first, second and third degrees. But modern designations (such as a 4, a 5) in the XVII were introduced by Rene Descartes.
Modern definitions and notation of degrees with zero, negative and fractional exponents originate from the works of the English mathematicians John Wallis (1616-1703) and Isaac Newton (1643-1727).
On the advisability of introducing zero, negative and fractional indicators and modern symbols first wrote in detail in 1665 by the English mathematician John Wallis. His business was completed by Isaac Newton, who began to systematically apply new symbols, after which they entered general use.
The introduction of a degree with a rational exponent is one of many examples of generalizing the concepts of mathematical action. A degree with zero, negative and fractional exponents is determined in such a way that the same rules of action are applied to it that take place for a degree with a natural exponent, i.e. so that the basic properties of the original definite concept of degree are preserved.
The new definition of a degree with a rational exponent does not contradict the old definition of a degree with a natural exponent, that is, the meaning of a new definition of a degree with a rational exponent is preserved for the particular case of a degree with a natural exponent. This principle, observed when generalizing mathematical concepts, is called the principle of permanence (preservation of constancy). It was expressed in an imperfect form in 1830 by the English mathematician J. Peacock; it was fully and clearly established by the German mathematician G. Hankel in 1867.
VIII. Check yourself.
Independent work by cards (answers open on the board) .
Option 1
1. Calculate: (1 point)
(a + 3a 1 \ 2): (a 1 \ 2 +3)
Option 2
1. Calculate: (1 point)
2. Simplify the expression: 1 point each
a) x 1.6 x 0.4 b) (x 3 \ 8) -5 \ 6
3. Solve the equation: (2 points)
4. Simplify the expression: (2 points)
5. Find the value of the expression: (3 points)
IX. Summing up the lesson.
What formulas and rules did you remember in the lesson?
Analyze your work in the lesson.
The work of students in the lesson is assessed.
NS. Homework... К: Р IV (repeat) Art. 156-157 No. 4 (a-c), No. 7 (a-c),
Optional: No. 16
Application
Evaluation paper
F / I / student __________________________________________
| Crossword | Warm up | Work in | Equations | Check yourself (s \ r) | ||
Card number 1
1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3
Decoder
Card number 2
1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3
Decoder
Card number 3
1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3
Decoder
Card number 1
1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3
Decoder
Card number 2
1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3
Decoder
Card number 3
1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3
Decoder
Card number 1
1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3
Decoder
Card number 2
1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3
Decoder
Card number 3
1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3
Decoder
| Option 1 1. Calculate: (1 point) 2. Simplify the expression: 1 point each a) x 1 \ 2 x 3 \ 4 b) (x -5 \ 6) -2 \ 3 c) x -1 \ 3: x 3 \ 4 d) (0.04x 7 \ 8) -1 \ 2 3. Solve the equation: (2 points) 4. Simplify the expression: (2 points) (a + 3a 1 \ 2): (a 1 \ 2 +3) 5. Find the value of the expression: (3 points) (Y 1 \ 2 -2) -1 - (Y 1 \ 2 +2) -1 at y = 18 | Option 2 1. Calculate: (1 point) 2. Simplify the expression: 1 point each a) x 1.6 x 0.4 b) (x 3 \ 8) -5 \ 6 c) x 3 \ 7: x -2 \ 3 d) (0.008x -6 \ 7) -1 \ 3 3. Solve the equation: (2 points) 4. Simplify the expression: (2 points) (in 1.5 s - sun 1.5): (in 0.5 - s 0.5) 5. Find the value of the expression: (3 points) (x 3 \ 2 + x 1 \ 2): (x 3 \ 2 -x 1 \ 2) at x = 0.75 |
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Municipal government educational institution
the main comprehensive school № 25
Algebra lesson
Topic:
« Converting expressions containing exponents with fractional exponents "
Developed by:
,
mathematic teacher
highest tovalidation category
Nodal
2013
Lesson topic: Convert expressions containing fractional exponents
The purpose of the lesson:
1. Further formation of skills, knowledge, skills of transforming expressions containing degrees with fractional indicators
2. Development of the ability to find errors, development of thinking, creativity, speech, computing skills
3. Education of independence, interest in the subject, attentiveness, accuracy.
TSO: a magnetic board, control cards, tables, individual cards, schoolchildren have blank signed sheets for individual work on the table, a crossword puzzle, tables for a mathematical warm-up, a multimedia projector.
Lesson type: securing the ZUN.
Lesson plan in time
1. Organizational moments (2 min)
2. Homework check (5 min)
3. Solving the crossword puzzle (3 min)
4. Math warm-up (5 min)
5. Solution of frontal strengthening exercises (7 min)
6. Individual work (10 min)
7. Repetition exercise solution (5 min)
8. Lesson summary (2 min)
9. Homework (1 min)
During the classes
1) Checking homework in the form of a peer review ... Good students check the notebooks of weak children. And the weak guys check the strong ones on the model of the control card. Homework is given in two versions.
I option the task is not difficult
II option the task is difficult
As a result of the check, the guys underline the mistakes with a simple pencil and give a grade. Finally, I check the work after the guys hand over their notebooks after the lesson. I ask the guys for the results of their verification and put marks for this type of work in my summary table.
2) A crossword puzzle is offered to check the theoretical material..
Vertically:
1. The property of multiplication used when multiplying a monomial by a polynomial?
2. The effect of exponents in raising a degree to an exponent?
3. A zero-score degree?
4. A product consisting of the same factors?
Horizontally:
5. Root n - th degree of a non-negative number?
6. The effect of exponents when multiplying degrees?
7. The effect of exponents when dividing degrees?
8. The number of all the same factors?
3) Math warm-up
a) perform the calculation and use the cipher to read the word hidden in the problem.
On the board in front of you is a table. The table in column 1 contains examples that need to be calculated.
Key to the table
491/2 | ||
27-1/3 | ||
4*81/3 | ||
5*25-1/2 | ||
7*82/3 | ||
(49/144)1/2 | 7/12 | |
(27*64)1/3 |
7/12 |
And write the answer in the column II, and in column III put the letter corresponding to this answer.
Teacher: So, the encrypted word "degree". In the next task, we work with the 2nd and 3rd degree
b) The game "Look not to be mistaken"
Put a number instead of dots
a) x = (x ...) 2; b) a3 / 2 = (a1 / 2) ...; c) a = (a1 / 3) ...; d) 5 ... = (51/4) 2; e) 34/3 = (34/9) ...; f) 74/5 = (7 ...) 2; g) x1 / 2 = (x ...) 2; h) y1 / 2 = (y ...) 2
Let's find the error:
A1 / 4 - 2a1 / 2 + 1 = (a1 /
So, guys, what needed to be applied to complete this task:
Property of degrees: when raising a degree to a power, the indicators are multiplied;
4) Now let's get down to the frontal writing work. using the results of previous work. Open notebooks, write down the number, the topic of the lesson.
№ 000
a) a - b = (a1 / 2) 2 - (b1 / 2) 2 = (a1 / 2 - b1 / 2) * (a1 / 2 + b1 / 2)
b) a - c = (a1 / 3) 3 - (b1 / 3) 3 = (a1 / 3 - b1 / 3) * (a2 / 3 + a1 / 3 b1 / 3 + b2 / 3)
No. 000 (a, c, d, e)
but ) m2 - 5 = m2 - (m1 / 2) 2 = (m - 51/2) * (m + 51/2)
c) a3 - 4 = (a3 / 2) 2 - 22 = (a3 / 2 - 2) * (a3 / 2 +2)
d) x2 / 5 - y4 / 5 = (x1 / 5) 2 - (y2 / 5) 2 = (x1 / 5 - y2 / 5) * (x1 / 5 + y2 / 5)
e) 4 - a = 22 - (a1 / 2) 2 = (2 - a1 / 2) * (2 + a1 / 2)
No. 000 (a, d, f)
a) x3 - 2 = x3 - (21/3) 3 = (x - 21/3) * (x2 + 21/3 x + 22/3)
d) a6 / 5 + 27 = (a2 / 5) 3 + 33 = (a2 / 5 + 3) * (a4 / 3 - 3 a2 / 5 + 9)
f) 4 + y = (41/3) 3 + (y1 / 3) 3 = (41/3 + y1 / 3) * (42/3 + 41/3 y1 / 3 + y2 / 3)
Grade
5) Work on individual cards in four options on separate sheets
Tasks with varying degrees of difficulty are performed without any teacher's advice.
I check the work right away and put the marks in my table and on the sheets of the guys.
No. 000 (a, c, d, h)
a) 4 * 31/2 / (31/2 - 3) = 4 * 31/2/31/2 * (1 - 31/2) = 4 / (1 - 31/2)
c) x + x1 / 2 / 2x = x1 / 2 * (x1 / 2 + 1) / 2 * (x1 / 2) 2 = (x1 / 2 + 1) / 2x1 / 2
e) (a2 / 3 - b2 / 3) / (a1 / 3 + b1 / 3) = (a1 / 3) 2 - (b1 / 3) 2 / (a1 / 3 + b1 / 3) = (a1 / 3 + b1 / 3) * (a1 / 3 - b1 / 3) / (a1 / 3 + b1 / 3) = a1 / 3 - b1 / 3
h) (x2 / 3 - x1 / 3 y1 / 3 + y2 / 3) / (x + y) = ((x1 / 3) 2 - x1 / 3 y1 / 3 + (y1 / 3) 2) / (( x1 / 3) 3 + (y1 / 3) 3) = ((x1 / 3) 2 - x1 / 3 y1 / 3 + (y1 / 3) 2) / (x1 / 3 + y1 / 3) * ((x1 / 3) 2 - x1 / 3 y1 / 3 + (y1 / 3) 2) = 1 / (x1 / 3 + y1 / 3)
7) Work on individual cards with varying degrees of difficulty... In some exercises there are teacher recommendations, since the material is complicated and weak children find it difficult to cope with the work
There are also four options. Assessment takes place immediately. I put all the grades in the table.
Problem number from the collection
The teacher asks questions:
1. What should be found in the problem?
2. What do you need to know for this?
3. How to express the time of 1 pedestrian and 2 pedestrians?
4. Compare the time of 1 and 2 pedestrians according to the condition of the problem and make an equation.
The solution of the problem:
Let x (km / h) be the speed of 1 pedestrian
X +1 (km / h) - speed of 2 pedestrians
4 / x (h) - pedestrian time
4 / (x +1) (h) - time of the second pedestrian
By the condition of the problem 4 / x> 4 / (x +1) for 12 min
12 minutes = 12/60 hours = 1/5 hours
We make the equation
X / 4 - 4 / (x +1) = 1/5
NOZ: 5x (x +1) ≠ 0
5 * 4 * (x + 1) - 5 * 4x = x * (x + 1)
20x + 20 - 20x - x2 - x = 0
X2 + x –20 = 0
D = 1 - 4 * (- 20) = 81.81> 0.2 k
х1 = (-1 -√81) / (- 2) = 5 km / h - speed of 1 pedestrian
x2 = (-1 + √81) / (- 2) = 4 - does not fit within the meaning of the problem, since x> 0
Answer: 5 km / h - speed of 2 pedestrians
9) Lesson summary: So, guys, today in the lesson we have consolidated knowledge, skills, skills in transforming expressions containing degrees, used formulas for abbreviated multiplication, taking the common factor out of the parentheses, and repeated the material covered. I point out the advantages and disadvantages.
Summing up the lesson in the table.
Crossword | Mat. warm-up | Front. Job | Ind. work K-1 | Ind. work K-2 | ||||
10) I announce the grades. Homework
Individual cards K - 1 and K - 2
I change B - 1 and B - 2; B - 3 and B - 4, since they are equivalent
Appendices to the lesson.
1) Homework cards
1.simplify
a) (x1 / 2 - y1 / 2) 2 + 2x1 / 2 y1 / 2
b) (a3 / 2 + 5a1 \ 2) 2 - 10a2
2.present as a sum
a) a1 / 3 c1 \ 4 * (b2 / 3 + c3 / 4)
b) (a1 / 2 - b1 / 2) * (a + a1 / 2 b1 \ 2 + c)
3.Pull out the common factor
c) 151/3 +201/3
1.simplify
a) √m + √n - (m1 / 4 - n1 / 4) 2
b) (a1 / 4 + b1 / 4) * (a1 / 8 + b1 / 8) * (a1 \ 8 - b1 / 8)
2.present as a sum
a) x0.5 y0.5 * (x-0.5 - y1.5)
b) (x1 / 3 + y1 / 3) * (x2 \ 3 - x1 / 3 y1 \ 3 + y2 / 3)
3. Take the common factor out of the parentheses
b) c1 \ 3 - c
c) (2а) 1/3 - (5а) 1/3
2) control card for B - 2
a) √m + √n - (m 1 | 4 - n 1 | 4) 2 = m 1 | 2 + n 1 | 2 - ((m 1 | 2) 2 - 2 m 1/4 n 1/4 + (n 1/2) 2) = m 1/2 + n 1/2 - m 1/2 + 2 m 1/4 n 1/4 - n 1/2 = 2 m 1/4 n 1/4
b) (a1 / 4 + b1 / 4) * (a1 / 8 + b1 / 8) * (a1 / 8 - b1 / 8) = (a1 / 4 + b1 / 4) * (a1 / 8) 2 - ( b1 / 8) 2 = (a1 / 4 + b1 / 4) * (a1 / 4 - b1 / 4) = (a1 / 4) 2 - (b1 / 4) 2 = a1 / 2 - b1 / 2
a) x0.5 y0.5 * (x-0.5-y1.5) = x0.5 y0.5 x-0.5 - x0.5 y0.5y1.5 = x0 y0.5 - x0.5 y2 = y0.5 - x0.5 y2
b) (x1 / 3 + y1 / 3) * (x2 / 3 - x1 / 3 y1 \ 3 + y2 / 3) = (x1 \ 3 + y1 / 3) * ((x1 / 3) 2 - x1 / 3 y1 \ 3 + (y1 / 3) 2) = (x1 / 3) 2 + (y1 / 3) 2 = x + y
a) 3 - 31/2 = 31/2 * (31/2 - 1)
b) в1 / 3 - в = в1 / 3 * (1 - в2 / 3)
c) (2a) 1/3 - (5a) 1/3 = a1 / 3 * (21/3 - 51/3)
3) Cards for the first individual work
a) a - y, x ≥ 0, y ≥ 0
b) a - u, a ≥ 0
1. Factor by presenting as the difference of squares
a) a1 / 2 - b1 / 2
2. Factor by representing as difference or sum of cubes
a) c1 / 3 + d1 / 3
1. Factor by presenting as the difference of squares
a) X1 / 2 + Y1 / 2
b) X1 / 4 - Y1 / 4
2. Factor by representing as difference or sum of cubes
4) cards for the second individual work
a) (x - x1 / 2) / (x1 / 2 - 1)
Indication: x1 / 2 put out the numerators from the parenthesis
b) (a - c) / (a1 / 2 - b1 / 2)
Hint: a - b = (a1 / 2) 2 - (b1 / 2) 2
Reduce the fraction
a) (21/4 - 2) / 5 * 21/4
Note: place 21/4 outside the parenthesis
b) (a - c) / (5a1 / 2 - 5v1 / 2)
Note: a - b = (a1 / 2) 2– (b1 / 2) 2
Option 3
1. Reduce the fraction
a) (x1 / 2 - x1 / 4) / x3 / 4
Note: x1 / 4 is out of the parenthesis
b) (а1 / 2 - в1 / 2) / (4а1 / 4 - 4в1 / 4)
Option 4
Reduce the fraction
a) 10 / (10 - 101/2)
b) (a - c) / (a2 / 3 + a1 \ 3b1 / 3 + B 1/3)
Expressions, expression conversion
Power expressions (expressions with powers) and their conversion
In this article, we will talk about converting power expressions. First, we will focus on transformations that are performed with expressions of any kind, including exponential expressions, such as expanding parentheses, casting similar terms. And then we will analyze the transformations inherent in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.
Page navigation.
What are exponential expressions?
The term "exponential expressions" is practically not found in school textbooks of mathematics, but it appears quite often in collections of problems, especially those intended for preparation for the exam and the exam, for example,. After analyzing the tasks in which you need to perform any actions with exponential expressions, it becomes clear that expres- sions are understood as expressions containing degrees in their records. Therefore, for yourself, you can accept the following definition:
Definition.
Power expressions Are expressions containing degrees.
Let us give examples of exponential expressions... Moreover, we will represent them according to how the development of views on occurs from a degree with a natural indicator to a degree with a real indicator.
As you know, first there is an acquaintance with the power of a number with a natural exponent, at this stage the first simplest power expressions of the type 3 2, 7 5 +1, (2 + 1) 5, (−0,1) 4, 3 a 2 −a + a 2, x 3−1, (a 2) 3, etc.
A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, 
, a −2 + 2 b −3 + c 2.
In high school, they return to degrees again. There, a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: 
, , 
etc. Finally, degrees with irrational indicators and expressions containing them are considered:,.
The matter is not limited to the listed power expressions: the variable penetrates further into the exponent, and, for example, such expressions 2 x 2 +1 or 
... And after meeting with, expressions with powers and logarithms begin to occur, for example, x 2 · lgx −5 · x lgx.
So, we figured out the question of what are exponential expressions. Next, we will learn to transform them.
Basic types of transformations of power expressions
With exponential expressions, you can perform any of the basic identical transformations of expressions. For example, you can expand parentheses, replace numeric expressions with their values, provide similar terms, etc. Naturally, in this case it is necessary to follow the accepted procedure for performing actions. Here are some examples.
Example.
Evaluate the value of the exponential expression 2 3 · (4 2 −12).
Solution.
According to the order of performing the actions, we first perform the actions in brackets. There, firstly, we replace the degree 4 2 with its value 16 (see if necessary), and secondly, we calculate the difference 16−12 = 4. We have 2 3 (4 2 −12) = 2 3 (16−12) = 2 3 4.
In the resulting expression, replace the power 2 3 with its value 8, after which we calculate the product 8 4 = 32. This is the desired value.
So, 2 3 (4 2 −12) = 2 3 (16−12) = 2 3 4 = 8 4 = 32.
Answer:
2 3 (4 2 −12) = 32.
Example.
Simplify Power Expressions 3 a 4 b −7 −1 + 2 a 4 b −7.
Solution.
Obviously, this expression contains similar terms 3 · a 4 · b −7 and 2 · a 4 · b −7, and we can bring them:.
Answer:
3 a 4 b −7 −1 + 2 a 4 b −7 = 5 a 4 b −7 −1.
Example.
Imagine an expression with powers as a product.
Solution.
To cope with the task, the representation of the number 9 in the form of a power of 3 2 and the subsequent use of the formula for abbreviated multiplication is the difference of squares: 
Answer:
There are also a number of identical transformations inherent in power expressions. Then we will analyze them.
Working with base and exponent
There are degrees, the base and / or exponent of which are not just numbers or variables, but some expressions. As an example, we present the entries (2 + 0.37) 5-3.7 and (a (a + 1) -a 2) 2 (x + 1).
When working with such expressions, you can replace both the expression based on the degree and the expression in the exponent with an identically equal expression on the ODZ of its variables. In other words, we can, according to the rules known to us, separately transform the base of the degree, and separately - the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.
Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the above exponential expression (2 + 0.3 · 7) 5-3.7, you can perform actions with the numbers in the base and exponent, which will allow you to go to the power 4.1 1.3. And after expanding the parentheses and reducing similar terms in the base of the degree (a (a + 1) −a 2) 2 (x + 1), we get a power expression of a simpler form a 2
Using degree properties
One of the main tools for converting expres- sions with powers is equalities reflecting. Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following power properties are valid:
- a r a s = a r + s;
- a r: a s = a r − s;
- (a b) r = a r b r;
- (a: b) r = a r: b r;
- (a r) s = a r s.
Note that for natural, integer, and also positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n, the equality a m a n = a m + n is true not only for positive a, but also for negative ones, and for a = 0.
At school, the main attention when transforming power expressions is focused precisely on the ability to choose a suitable property and apply it correctly. In this case, the bases of degrees are usually positive, which allows using the properties of degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of admissible values of variables is usually such that the bases take only positive values on it, which allows you to freely use the properties of degrees. In general, you need to constantly ask yourself whether it is possible in this case to apply any property of degrees, because inaccurate use of properties can lead to a narrowing of the ODV and other troubles. These points are discussed in detail and with examples in the article on conversion of expressions using degree properties. Here we restrict ourselves to a few simple examples.
Example.
Imagine the expression a 2.5 · (a 2) −3: a −5.5 as a power with base a.
Solution.
First, we transform the second factor (a 2) −3 by the property of raising a power to a power: (a 2) −3 = a 2 (−3) = a −6... The original exponential expression will then take the form a 2.5 · a −6: a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 a -6: a -5.5 =
a 2.5-6: a -5.5 = a -3.5: a -5.5 =
a −3.5 - (- 5.5) = a 2.
Answer:
a 2.5 (a 2) −3: a −5.5 = a 2.
Power properties are used both from left to right and from right to left when transforming exponential expressions.
Example.
Find the value of the exponential expression.
Solution.
Equality (a b) r = a r b r, applied from right to left, allows you to go from the original expression to the product of the form and further. And when multiplying degrees with the same bases, the indicators add up: 
.
It was possible to perform the transformation of the original expression in another way: 
Answer:
.
Example.
Given the exponential expression a 1.5 −a 0.5 −6, enter the new variable t = a 0.5.
Solution.
The degree a 1.5 can be represented as a 0.5 · 3 and further, based on the property of the degree to the degree (a r) s = a r · s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6 = (a 0.5) 3 −a 0.5 −6... Now it is easy to introduce a new variable t = a 0.5, we get t 3 −t − 6.
Answer:
t 3 −t − 6.
Converting fractions containing powers
Power expressions can contain fractions with powers or be such fractions. Any of the basic transformations of fractions that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain powers can be canceled, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate the words spoken, consider the solutions of several examples.
Example.
Simplify exponential expression 
.
Solution.
This exponential expression is a fraction. Let's work with its numerator and denominator. In the numerator, we open the brackets and simplify the expression obtained after that using the properties of the powers, and in the denominator we give similar terms: 
And we also change the sign of the denominator by placing a minus in front of the fraction: 
.
Answer:
.
The reduction of fractions containing powers to a new denominator is carried out similarly to the reduction of rational fractions to a new denominator. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that the reduction to a new denominator can lead to a narrowing of the ODV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values of the variables from the ODZ variables for the original expression.
Example.
Reduce fractions to a new denominator: a) to the denominator a, b) 
to the denominator.
Solution.
a) In this case, it is quite easy to figure out which additional factor helps to achieve the desired result. This is a factor of a 0.3, since a 0.7 · a 0.3 = a 0.7 + 0.3 = a. Note that on the range of permissible values of the variable a (this is the set of all positive real numbers) the degree a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of the given fraction by this additional factor: 
b) Looking more closely at the denominator, you can find that 
and multiplying this expression by will give the sum of cubes and, that is,. And this is the new denominator to which we need to reduce the original fraction.
This is how we found an additional factor. On the range of valid values of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it: 
Answer:
but) 
, b) 
.
The reduction of fractions containing powers is also nothing new: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are canceled.
Example.
Reduce the fraction: a) 
, b).
Solution.
a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which is 15. Also, obviously, one can perform a reduction by x 0.5 +1 and by 
... Here's what we have: 
b) In this case, the same factors in the numerator and denominator are not immediately visible. To get them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator into factors according to the formula for the difference of squares: 
Answer:
but) 
b) 
.
Reducing fractions to a new denominator and reducing fractions is mainly used to perform actions with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are brought to a common denominator, after which the numerators are added (subtracted), and the denominator remains the same. The result is a fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by the inverse of the fraction.
Example.
Follow the steps 
.
Solution.
First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is 
, after which we subtract the numerators: 
Now we multiply the fractions: 
Obviously, it is possible to cancel by a power of x 1/2, after which we have 
.
You can also simplify the exponential expression in the denominator by using the difference of squares formula: 
.
Answer:
Example.
Simplify exponential expression 
.
Solution.
Obviously, this fraction can be canceled by (x 2.7 +1) 2, this gives the fraction 
... It is clear that something else needs to be done with the degrees of x. To do this, we transform the resulting fraction into a product. This gives us the opportunity to use the property of dividing degrees with the same bases: 
... And at the end of the process, we pass from the last product to a fraction.
Answer:
.
And we also add that it is possible and in many cases desirable to transfer multipliers with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, an exponential expression can be replaced with.
Converting expressions with roots and powers
Often in expressions in which some transformations are required, along with powers with fractional exponents, roots are also present. To transform such an expression to the desired form, in most cases it is enough to go only to the roots or only to the powers. But since it is more convenient to work with degrees, they usually go from roots to degrees. However, it is advisable to carry out such a transition when the ODV of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODV into several intervals (we discussed in detail in the article the transition from roots to powers and back. a degree with an irrational indicator is introduced, which makes it possible to talk about a degree with an arbitrary real indicator. exponential function, which is analytically set by the degree, at the base of which is the number, and in the indicator - the variable. So we are faced with exponential expressions containing numbers in the base of the degree, and in the exponent - expressions with variables, and naturally there is a need to perform transformations of such expressions.
It should be said that the transformation of expressions of this type usually has to be performed when solving exponential equations and exponential inequalities and these conversions are pretty simple. In the overwhelming majority of cases, they are based on the properties of the degree and are mainly aimed at introducing a new variable in the future. We can demonstrate them by the equation 5 2 x + 1 −3 5 x 7 x −14 7 2 x − 1 = 0.
First, the degrees in which the sum of a variable (or expressions with variables) and a number is found are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 = 0,
5 5 2 x −3 5 x 7 x −2 7 2 x = 0.
Further, both sides of the equality are divided by the expression 7 2 x, which takes only positive values on the ODZ of the variable x for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on the subsequent transformations of expressions with powers ): 
Fractions with powers are now canceled, which gives 
.
Finally, the ratio of degrees with the same exponents is replaced by the degrees of relations, which leads to the equation 
which is equivalent 
... The performed transformations allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation
The arithmetic action that is performed last when calculating the value of the expression is the "main" one.
That is, if you substitute any (any) numbers instead of letters, and try to calculate the value of the expression, then if the last action is multiplication, then we have a product (the expression is factorized).
If the last action is addition or subtraction, this means that the expression is not factorized (and therefore cannot be canceled).
To fix the solution yourself, take a few examples:
Examples:
Solutions:
1. I hope you didn't rush to cut u right away? It was still not enough to "cut" units like this:
The first action should be factoring:
4. Addition and subtraction of fractions. Bringing fractions to a common denominator.
Adding and subtracting ordinary fractions is a very familiar operation: we look for a common denominator, multiply each fraction by the missing factor and add / subtract the numerators.
Let's remember:
Answers:
1. The denominators and are mutually prime, that is, they have no common factors. Therefore, the LCM of these numbers is equal to their product. This will be the common denominator:
2. Here the common denominator is:
3. First thing here mixed fractions we turn them into wrong ones, and then - according to the usual scheme:
It is completely different if the fractions contain letters, for example:
Let's start simple:
a) Denominators do not contain letters
Here everything is the same as with ordinary number fractions: find the common denominator, multiply each fraction by the missing factor and add / subtract the numerators:
now in the numerator you can bring similar ones, if any, and decompose into factors:
Try it yourself:
Answers:
b) Denominators contain letters
Let's remember the principle of finding a common denominator without letters:
· First of all, we determine the common factors;
· Then write out all common factors one time;
· And multiply them by all other factors that are not common.
To determine the common factors of the denominators, we first decompose them into prime factors:
Let's emphasize the common factors:
Now let's write out the common factors one time and add to them all non-common (not underlined) factors:
This is the common denominator.
Let's go back to the letters. The denominators are shown in exactly the same way:
· We decompose the denominators into factors;
· Determine common (identical) factors;
· Write out all common factors once;
· We multiply them by all other factors, not common.
So, in order:
1) we decompose the denominators into factors:
2) we determine the common (identical) factors:
3) we write out all the common factors one time and multiply them by all the other (unstressed) factors:
So the common denominator is here. The first fraction must be multiplied by, the second by:
By the way, there is one trick:
For example: .
We see the same factors in the denominators, only all with different indicators. The common denominator will be:
to the extent
to the extent
to the extent
in degree.
Let's complicate the task:
How do you make fractions the same denominator?
Let's remember the basic property of a fraction:
Nowhere is it said that the same number can be subtracted (or added) from the numerator and denominator of a fraction. Because this is not true!
See for yourself: take any fraction, for example, and add some number to the numerator and denominator, for example,. What has been learned?
So, another unshakable rule:
When bringing fractions to a common denominator, use only multiplication!
But what must be multiplied by in order to receive?
Here on and multiply. And multiply by:
Expressions that cannot be factorized will be called “elementary factors”.
For example, is an elementary factor. - too. But - no: it is factorized.
What do you think about expression? Is it elementary?
No, since it can be factorized:
(you already read about factorization in the topic "").
So, the elementary factors into which you expand the expression with letters are analogous to the prime factors into which you expand the numbers. And we will deal with them in the same way.
We see that there is a factor in both denominators. It will go to the common denominator in power (remember why?).
The factor is elementary, and it is not common for them, which means that the first fraction will simply have to be multiplied by it:
Another example:
Solution:
Before multiplying these denominators in a panic, you need to think about how to factor them? They both represent:
Excellent! Then:
Another example:
Solution:
As usual, factor the denominators. In the first denominator, we simply put it outside the brackets; in the second - the difference of squares:
It would seem that there are no common factors. But if you look closely, then they are so similar ... And the truth:
So we will write:
That is, it turned out like this: inside the parenthesis, we swapped the terms, and at the same time the sign in front of the fraction changed to the opposite. Take note, you will have to do this often.
Now we bring to a common denominator:
Got it? Let's check it out now.
Tasks for independent solution:
Answers:
Here we must remember one more - the difference between the cubes:
Please note that the denominator of the second fraction is not the "square of the sum" formula! The square of the sum would look like this:.
A is the so-called incomplete square of the sum: the second term in it is the product of the first and the last, and not their doubled product. The incomplete square of the sum is one of the factors in the expansion of the difference of cubes:
What if there are already three fractions?
Yes, the same! First of all, let's make sure that the maximum number of factors in the denominators is the same:
Pay attention: if you change the signs within one parenthesis, the sign in front of the fraction changes to the opposite. When we change the signs in the second parenthesis, the sign in front of the fraction is reversed again. As a result, it (the sign in front of the fraction) has not changed.
We write out the first denominator completely into the common denominator, and then add to it all the factors that have not yet been written, from the second, and then from the third (and so on, if there are more fractions). That is, it turns out like this:
Hmm ... With fractions, it's clear what to do. But what about the deuce?
It's simple: you can add fractions, right? So, we need to make sure that the deuce becomes a fraction! Remember: a fraction is a division operation (the numerator is divided by the denominator, in case you suddenly forgot). And there is nothing easier than dividing a number by. In this case, the number itself will not change, but it will turn into a fraction:
Exactly what is needed!
5. Multiplication and division of fractions.
Well, the hardest part is over now. And ahead of us is the simplest, but at the same time the most important:
Procedure
What is the procedure for calculating a numeric expression? Remember by counting the meaning of such an expression:
Have you counted?
It should work.
So, I remind you.
The first step is to calculate the degree.
The second is multiplication and division. If there are several multiplications and divisions at the same time, they can be done in any order.
And finally, we do addition and subtraction. Again, in any order.
But: the expression in parentheses is evaluated out of order!
If several brackets are multiplied or divided by each other, we first calculate the expression in each of the brackets, and then we multiply or divide them.
What if there are more brackets inside the brackets? Well, let's think about it: some expression is written inside the brackets. And when evaluating an expression, what is the first thing to do? That's right, calculate the parentheses. Well, we figured it out: first we calculate the inner brackets, then everything else.
So, the procedure for the expression above is as follows (the current action is highlighted in red, that is, the action that I am performing right now):
Okay, it's all simple.
But this is not the same as an expression with letters?
No, it's the same! Only, instead of arithmetic operations, you need to do algebraic ones, that is, the actions described in the previous section: bringing similar, addition of fractions, reduction of fractions, and so on. The only difference is the effect of factoring polynomials (we often use it when working with fractions). Most often, for factoring, you need to use i or just put the common factor outside the parentheses.
Usually our goal is to present an expression in the form of a work or a particular.
For example:
Let's simplify the expression.
1) The first is to simplify the expression in parentheses. There we have the difference of fractions, and our goal is to present it as a product or quotient. So, we bring the fractions to a common denominator and add:
It is impossible to simplify this expression anymore, all the factors here are elementary (do you still remember what this means?).
2) We get:
Multiplication of fractions: what could be easier.
3) Now you can shorten:
That's it. Nothing complicated, right?
Another example:
Simplify the expression.
First try to solve it yourself, and only then see the solution.
Solution:
First of all, let's define the order of actions.
First, we add the fractions in brackets, we get one instead of two fractions.
Then we will divide the fractions. Well, add the result with the last fraction.
I will enumerate the actions schematically:
Now I will show the whole process, coloring the current action in red:
1. If there are similar ones, they must be brought immediately. At whatever moment we have similar ones, it is advisable to bring them right away.
2. The same applies to the reduction of fractions: as soon as there is an opportunity to reduce, it must be used. The exception is fractions that you add or subtract: if they now have the same denominators, then the reduction should be left for later.
Here are some tasks for you to solve on your own:
And promised at the very beginning:
Answers:
Solutions (concise):
If you have coped with at least the first three examples, then you have mastered the topic.
Now forward to learning!
TRANSFORMATION OF EXPRESSIONS. SUMMARY AND BASIC FORMULAS
Basic simplification operations:
- Bringing similar: to add (bring) such terms, you need to add their coefficients and assign the letter part.
- Factorization: factoring out the common factor, application, etc.
- Fraction reduction: the numerator and denominator of a fraction can be multiplied or divided by the same non-zero number, which does not change the value of the fraction.
1) numerator and denominator factor
2) if there are common factors in the numerator and denominator, they can be crossed out.IMPORTANT: only multipliers can be reduced!
- Addition and subtraction of fractions:
; - Multiplication and division of fractions:
;
