Lesson #30 (Algebra and the Beginnings of Analysis, Grade 11)
Lesson topic: Degree with a rational exponent.
Lesson goal: 1 . Expand the concept of degree, give the concept of degree with a rational indicator; to teach how to translate a degree with a rational indicator to the root and vice versa; calculate powers with a rational exponent.
2. Development of memory, thinking.
3. Formation of activity.
"Let someone try to cross out
from a mathematics degree and he will see
You won't get far without them." M.V. Lomonosov
During the classes.
I. Communication of the topic and purpose of the lesson.
II. Repetition and consolidation of the material covered.
1. Analysis of unsolved home examples.
2. Controlling independent work:
Option 1.
1. Solve the equation: √(2x - 1) = 3x - 12
2. Solve the inequality: √(3x - 2) ≥ 4 - x
Option 2.
1. Solve the equation: 3 - 2x \u003d √ (7x + 32)
2. Solve the inequality: √(3x + 1) ≥ x - 1
III. Learning new material.
1 . Recall the extension of the concept of numbers: N є Z є Q є R.
This is best represented as the diagram below:
Natural (N)
Zero
Non-negative numbers
Negative numbers
Fractional numbers
Integers (Z)
Irrational
Rational (Q)
Real numbers
2. In the lower grades, the concept of the degree of a number with an integer exponent was defined. a) Recall the definition of the degree a) with a natural, b) with a negative integer, c) with a zero exponent.Emphasize that the expression a n makes sense for all integers n and any values of a, except for a=0 and n≤0.
b) List the properties of degrees with an integer exponent.
3 . oral work.
one). Calculate: 1 -5 ; 4-3; (-100 ; (-5) -2 ; (1/2) -4 ; (3/7) -1 .
2). Write as a negative exponent:
1/4 5 ;1/21 3 ; 1/x 7 ; 1/a 9 .
3).Compare with unit: 12-3 ; 21 0 ; (0,6) -5 ; (5/19) -4 .
4 . Now you need to understand the meaning of expressions 3 0,4 ; 4 5/7 ; 5 -1/2 etc. To do this, it is necessary to generalize the concept of a degree in such a way that all the listed properties of degrees are satisfied. Consider the equality (a m/n ) n = a m . Then, by the definition of the nth root, it is reasonable to assume that a m/n will be the nth root of a m . The definition of degree with a rational exponent is given.
5. Consider examples 1 and 2 from the textbook.
6. Let us make a number of remarks related to the concept of a degree with a rational exponent.
Remark 1 : For any a>0 and rational number r, the number a r>0
Remark 2 : By the basic property of fractions, a rational number m/n can be written as mk/nk for any natural number k. Thenthe value of the degree does not depend on the form of writing a rational number, since a mk/nk = = nk √ a mk = n √ a m = a m/n
Note 3: When a Let's explain this with an example. Consider (-64) 1/3 = 3 √-64 = -4. On the other hand: 1/3 = 2/6 and then (-64) 1/ 3 = (-64) 2/6 = 6 √(-64) 2 = 6√64 2 = 6 √4 6 = 4. We get a contradiction.
Mathematics teacher: Nashkenova A.N. Maybalyk secondary school Outline of the lesson on the topic "Degree with a rational indicator"
(algebra, grade 11)
Lesson Objectives:
To expand and deepen students' knowledge of the degree of number; familiarization of students with the concept of degree with a rational indicator and their properties;
Develop knowledge, skills and abilities to calculate the values of expressions by using properties;
Continue work on developing the skills to analyze, compare, highlight the main thing, define and explain concepts;
To form communicative competencies, the ability to argue their actions, to cultivate independence, diligence.
Equipment: textbook, handout cards, laptop,presentation material power point ;
Lesson type: lesson of studying and primary consolidation of new knowledge.
Lesson plan:
1.Org. moment. - 1 minute.
2.Motivation of the lesson.-2 minutes
3. Actualization of basic knowledge. - 5 minutes.
4. Study of new material. - 15 minutes.
5. Physical education minute - 1 min.
6. Primary consolidation of the studied material - 10 min
7. Independent work. - 7 min.
8. Homework. - 2 minutes.
9. Reflection - 1 min.
10. The result of the lesson. - 1 minute.
During the classes
1. Organizational moment
Emotional mood for the lesson.
I want to work, I want
work,
I wish you success today.
After all, in the future all this is for you
come in handy.
And it will be easier for you in the future
to study(Slide #1)
2. Lesson motivation
The operations of raising to a power and extracting a root, like the four arithmetic operations, appeared as a result of a practical need. So, along with the task of calculating the area of a square, the sidea which is known, there was an inverse problem: “What length should the side of the square have so that its area is equal toin. In the 14-15 centuries, banks appeared in Western Europe, which gave money at interest to princes and merchants, financed long-distance travels and conquests at high interest rates. To facilitate the calculation of compound interest, we compiled tables by which you could immediately find out how much you need to pay throughP years, if the amount was borroweda onR % per annum. The amount paid is expressed by the formula: s = a(1 + ) P .Sometimes money was borrowed not for an integer number of years, but for example, for 2 years 6 months. If after 2.5 years the amounta apply to aq , then in the next 2.5 years it will increase by anotherq times and becomes equalaq 2 . After 5 years:a=(1 + 5 , that's why q 2 = (1 + 5 and means q =
(Slide 2) .
Thus, the idea of a degree with a fractional exponent was born.
3. Actualization of basic knowledge.
Questions:
1. What does the record mean;a P
2. What is a ?
3. What is P ?
4. a -P =?
5. Write down in your notebook the properties of the degree with an integer indicator.
6. What numbers are natural, whole, rational? Draw them using Euler circles.(Slide 3)
Answers: 1. Degree with an integer exponent
2. a- base
3. P- exponent
4. a -P =
5. Degree properties with integer exponent:
a m *a n = a (m+n) ;
a m : a n = a (m-n) ( at a not equal to zero );
(a m ) n = a (m*n) ;
(a*b) n = a n *b n ;
(a/b) n = (a n )/(b n ) (at b not equal to zero);
a 1 = a;
a 0 = 1 (when a not equal to zero);
These properties will be valid for any numbers a, b and any integers m and n.
6.1,2,3, …- positive numbers – set of natural numbers –N
0,-1,-2,-3,.. the number O and negative numbers - a set of integers -Z
Q , – fractional numbers (negative and positive) – set of rational numbers -Q ZN
Euler circles (slide 4)
4. Learning new material.
Let. a - non-negative number and you want to raise it to a fractional power . Do you know the equationa m ) n = a m n (slide 4) , i.e. the rule for raising a power to a power. In the above equation, suppose that m = , then we get: (a ) P = a =a (slide 4)
From this it can be concluded that isa root P - th degree from the numbera , i.e. a = . it follows that (a P ) = P =a (slide 4).
Consequently a =(a ) m =(a m ) = m . ( slide 4 ).
Thus, the following equality holds:a = m (slide 4)
Definition: degree of a non-negative number a with a rational , where - an irreducible fraction, the value of the root of the n-th degree from a number is called a t .
Therefore, by definition a = m (slide 5)
Let's look at example 1 : Write the exponent with a rational exponent as the nth root:
1)5 2)3,7 -0,7 3) ( ) (slide 6) Solution: 1) 5 = 2 = 2) 3,7 -0,7 = -7 3) ( ) = ( slide 7) Multiplication, division, exponentiation, and root extraction can be performed on powers with a rational exponent according to the same rules as with powers with integer exponents and powers with the same bases:a = a + a = a - (a ) = a * (a*b) = a * in ) = a / in where p, q are natural numbers, m, p are integers. (slide 8) 5. Physical education minuteTurn your gaze to the right
Turn your gaze to the left
Looked at the ceiling
We all looked ahead.
One - bend - unbend,
Two ─ bend - stretch
Three - in the hands of three claps,
Three head nods.
Five and six quietly sit down.
And on the road again! (slide 9)
6. Primary consolidation of the studied material:
Page 51, No. 90, No. 91 - complete in a notebook yourself,
with board check
7. Independent work
Option 1
(Slide 10)
Option 1
(Slide 11)
Perform independent work with peer review.
Answers:
Option 1
(Slide 12)
So, today in the lesson we got acquainted with the concept of a degree with a rational exponent and learned how to write it in the form of roots, apply the basic properties of degrees when finding the values of numerical expressions.8. Homework: No. 92, No. 93 Homework Information
9. Reflection
(Slide 13)
10. Lesson summary:
What is the similarity and difference between the degree with an integer indicator and the degree with a fractional indicator? (similarity: all properties of a degree with an integer exponent also hold for a degree with a rational exponent;
difference: degrees)
List degree properties with rational exponent
Lesson completed today
You can't find friends.
But everyone should know:
Knowledge, perseverance, work
Lead to progress in life.
Thank you for the lesson!
(slide 14)
Expressions, expression conversion
Power expressions (expressions with powers) and their transformation
In this article, we will talk about transforming expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening brackets, reducing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.
Page navigation.
What are Power Expressions?
The term "power expressions" is practically not found in school textbooks of mathematics, but it often appears in collections of problems, especially designed to prepare for the Unified State Examination and the OGE, for example,. After analyzing tasks in which it is required to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing degrees in their entries. Therefore, for yourself, you can take the following definition:
Definition.
Power expressions are expressions containing powers.
Let's bring examples of power expressions. Moreover, we will represent them according to how the development of views on from a degree with a natural indicator to a degree with a real indicator takes place.
As you know, first you get acquainted with the degree 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 the senior classes, they return to the degrees again. There, a degree with a rational exponent is introduced, which leads to the appearance of the corresponding power expressions: 
, , 
etc. Finally, degrees with irrational exponents and expressions containing them are considered: , .
The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and there are, for example, such expressions 2 x 2 +1 or 
. And after getting acquainted with, expressions with powers and logarithms begin to appear, for example, x 2 lgx −5 x lgx.
So, we figured out the question of what are power expressions. Next, we will learn how to transform them.
The main types of transformations of power expressions
With power expressions, you can perform any of the basic identical transformations of expressions. For example, you can expand brackets, replace numeric expressions with their values, add like terms, and so on. Naturally, in this case it is necessary to comply with the accepted order of actions. Let's give examples.
Example.
Calculate the value of the power expression 2 3 ·(4 2 −12) .
Solution.
According to the order of the actions, we first perform the actions in brackets. There, firstly, we replace the power of 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, we replace the power of 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 like terms 3 a 4 b −7 and 2 a 4 b −7 , and we can reduce them: .
Answer:
3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.
Example.
Express an expression with powers as a product.
Solution.
To cope with the task allows the representation of the number 9 as a power of 3 2 and subsequent use abbreviated multiplication formulas difference of squares: 
Answer:
There are also a number of identical transformations inherent in power expressions. Next, we will analyze them.
Working with base and exponent
There are degrees, in the basis and / or indicator of which are not just numbers or variables, but some expressions. As an example, let's write (2+0.3 7) 5−3.7 and (a (a+1)−a 2) 2 (x+1) .
When working with similar expressions, both the expression in the base of the degree and the expression in the exponent can be replaced by an identically equal expression on ODZ his variables. In other words, according to the rules known to us, we can separately convert the base of the degree, and separately - the indicator. It is clear that as a result of this transformation, an expression is 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 power expression (2+0.3 7) 5−3.7 mentioned above, you can perform operations with numbers in the base and exponent, which will allow you to go to the power of 4.1 1.3. And after opening the brackets and bringing like 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·(x+1) .
Using Power Properties
One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following power properties hold:
- 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 positive exponents, 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 in the transformation of power expressions is focused precisely on the ability to choose the appropriate property and apply it correctly. In this case, the bases of the degrees are usually positive, which allows you to use the properties of the degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of acceptable 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 to apply any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the ODZ and other troubles. These points are discussed in detail and with examples in the article. transformation of expressions using the properties of degrees. Here we confine ourselves to a few simple examples.
Example.
Express 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. In this case, the initial power expression will 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 \u003d a 2.
Power properties are used when transforming power expressions both from left to right and from right to left.
Example.
Find the value of the power 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 powers with the same base, the indicators add up: 
.
It was possible to perform the transformation of the original expression in another way: 
Answer:
.
Example.
Given a power expression a 1.5 −a 0.5 −6 , enter a new variable t=a 0.5 .
Solution.
The degree a 1.5 can be represented as a 0.5 3 and further on the basis of the property of the degree in the degree (a r) s =a r s applied from right to left, convert it to the form (a 0.5) 3 . In this way, 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 represent such fractions. To such fractions, any of the main fraction conversions, which are inherent in fractions of any kind. That is, fractions that contain degrees can be reduced, reduced to a new denominator, work separately with their numerator and separately with the denominator, etc. To illustrate the above words, consider the solutions of several examples.
Example.
Simplify Power Expression 
.
Solution.
This power 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 powers, and in the denominator we present similar terms: 
And we also change the sign of the denominator by placing a minus in front of the fraction: 
.
Answer:
.
Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. At the same time, 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 reduction to a new denominator can lead to a narrowing of the DPV. 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.
Bring the 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 what additional factor helps to achieve the desired result. This is a multiplier a 0.3, since a 0.7 a 0.3 = a 0.7+0.3 = a . Note that on the range of acceptable 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, we 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 bring the original fraction.
So we found an additional factor . The expression does not vanish on the range of acceptable values of the variables x and y, therefore, we can multiply the numerator and denominator of the fraction by it: 
Answer:
a) 
, b) 
.
There is also nothing new in the reduction of fractions containing degrees: the numerator and denominator are represented as a certain number of factors, and the same factors of the numerator and denominator are reduced.
Example.
Reduce the fraction: a) 
, b).
Solution.
a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which equals 15. Also, obviously, you can reduce 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 have to perform preliminary transformations. In this case, they consist in decomposing the denominator into factors according to the difference of squares formula: 
Answer:
a) 
b) 
.
Reducing fractions to a new denominator and reducing fractions is mainly used to perform operations on fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), and the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its reciprocal.
Example.
Follow the steps 
.
Solution.
First, we subtract the fractions in brackets. To do this, we bring them to a common denominator, which is 
, then subtract the numerators: 
Now we multiply fractions: 
Obviously, a reduction by the power x 1/2 is possible, after which we have 
.
You can also simplify the power expression in the denominator by using the difference of squares formula: 
.
Answer:
Example.
Simplify Power Expression 
.
Solution.
Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction 
. It is clear that something else needs to be done with the powers of x. To do this, we convert the resulting fraction into a product. This gives us the opportunity to use the property of dividing powers with the same bases: 
. And at the end of the process, we pass from the last product to the fraction.
Answer:
.
And we add that it is possible and in many cases desirable to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator by changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .
Converting expressions with roots and powers
Often in expressions in which some transformations are required, along with degrees with fractional exponents, there are also roots. To convert such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with degrees, they usually move from roots to degrees. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with degrees without the need to access the module or split the ODZ into several intervals (we discussed this in detail in the article, the transition from roots to powers and vice versa After getting acquainted with the degree with a rational exponent a degree with an irrational indicator is introduced, which makes it possible to speak of a degree with an arbitrary real indicator.At this stage, the school begins to study exponential function, which is analytically given by the degree, in the basis of which there is a number, and in the indicator - a variable. So we are faced with power expressions containing numbers in the base of the degree, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.
It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations and exponential inequalities, and these transformations are quite simple. In the vast majority of cases, they are based on the properties of the degree and are aimed mostly at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.
First, the exponents, in whose exponents the sum of some variable (or expression 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.
Next, both sides of the equality are divided by the expression 7 2 x , which takes only positive values on the ODZ 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 subsequent transformations of expressions with powers ): 
Now fractions with powers are cancelled, which gives 
.
Finally, the ratio of powers with the same exponents is replaced by powers of ratios, which leads to the equation 
, which is equivalent to 
. The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation
