Lesson and presentation on the topic: "Properties of the root of the nth degree. Theorems"
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Properties of the root of the nth degree. Theorems
Guys, we continue to study the roots of the nth degree of a real number. Like almost all mathematical objects, the roots of the nth degree have some properties, today we will study them.All the properties that we consider are formulated and proved only for non-negative values of the variables contained under the root sign.
In the case of an odd root exponent, they also hold for negative variables.
Theorem 1. The nth root of the product of two non-negative numbers is equal to the product of the nth roots of these numbers: $\sqrt[n](a*b)=\sqrt[n](a)*\sqrt[n]( b) $ .
Let's prove the theorem.
Proof. Guys, to prove the theorem, let's introduce new variables, denote:
$\sqrt[n](a*b)=x$.
$\sqrt[n](a)=y$.
$\sqrt[n](b)=z$.
We need to prove that $x=y*z$.
Note that the following identities also hold:
$a*b=x^n$.
$a=y^n$.
$b=z^n$.
Then the following identity also holds: $x^n=y^n*z^n=(y*z)^n$.
The degrees of two non-negative numbers and their exponents are equal, then the bases of the degrees themselves are equal. Hence $x=y*z$, which is what was required to be proved.
Theorem 2. If $a≥0$, $b>0$ and n is a natural number greater than 1, then the following equality holds: $\sqrt[n](\frac(a)(b))=\frac(\sqrt[ n](a))(\sqrt[n](b))$.
That is, the nth root of the quotient is equal to the quotient of the nth roots.
Proof.
To prove this, we use a simplified scheme in the form of a table:
Examples of calculating the nth root
Example.Calculate: $\sqrt(16*81*256)$.
Solution. Let's use Theorem 1: $\sqrt(16*81*256)=\sqrt(16)*\sqrt(81)*\sqrt(256)=2*3*4=24$.
Example.
Calculate: $\sqrt(7\frac(19)(32))$.
Solution. Let's represent the radical expression as an improper fraction: $7\frac(19)(32)=\frac(7*32+19)(32)=\frac(243)(32)$.
Let's use Theorem 2: $\sqrt(\frac(243)(32))=\frac(\sqrt(243))(\sqrt(32))=\frac(3)(2)=1\frac(1) (2)$.
Example.
Calculate:
a) $\sqrt(24)*\sqrt(54)$.
b) $\frac(\sqrt(256))(\sqrt(4))$.
Solution:
a) $\sqrt(24)*\sqrt(54)=\sqrt(24*54)=\sqrt(8*3*2*27)=\sqrt(16*81)=\sqrt(16)*\ sqrt(81)=2*3=6$.
b) $\frac(\sqrt(256))(\sqrt(4))=\sqrt(\frac(256)(4))=\sqrt(64)=24$.
Theorem 3. If $a≥0$, k and n are natural numbers greater than 1, then the equality is true: $(\sqrt[n](a))^k=\sqrt[n](a^k)$.
To raise a root to a natural power, it is enough to raise the radical expression to this power.
Proof.
Let's consider a special case for $k=3$. Let's use Theorem 1.
$(\sqrt[n](a))^k=\sqrt[n](a)*\sqrt[n](a)*\sqrt[n](a)=\sqrt[n](a*a *a)=\sqrt[n](a^3)$.
The same can be proved for any other case. Guys, prove it yourself for the case when $k=4$ and $k=6$.
Theorem 4. If $a≥0$ b n,k are natural numbers greater than 1, then the equality is true: $\sqrt[n](\sqrt[k](a))=\sqrt(a)$.
To extract a root from a root, it is enough to multiply the exponents of the roots.
Proof.
Let us prove again briefly using the table. To prove this, we use a simplified scheme in the form of a table: 
Example.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.
$\sqrt(\sqrt(a))=\sqrt(a)$.
Theorem 5. If the indices of the root and the root expression are multiplied by the same natural number, then the value of the root will not change: $\sqrt(a^(kp))=\sqrt[n](a)$.
Proof.
The principle of the proof of our theorem is the same as in other examples. Let's introduce new variables:
$\sqrt(a^(k*p))=x=>a^(k*p)=x^(n*p)$ (by definition).
$\sqrt[n](a^k)=y=>y^n=a^k$ (by definition).
We raise the last equality to the power p
$(y^n)^p=y^(n*p)=(a^k)^p=a^(k*p)$.
Got:
$y^(n*p)=a^(k*p)=x^(n*p)=>x=y$.
That is, $\sqrt(a^(k*p))=\sqrt[n](a^k)$, which was to be proved.
Examples:
$\sqrt(a^5)=\sqrt(a)$ (divided by 5).
$\sqrt(a^(22))=\sqrt(a^(11))$ (divided by 2).
$\sqrt(a^4)=\sqrt(a^(12))$ (multiplied by 3).
Example.
Run actions: $\sqrt(a)*\sqrt(a)$.
Solution.
The exponents of the roots are different numbers, so we cannot use Theorem 1, but by applying Theorem 5 we can get equal exponents.
$\sqrt(a)=\sqrt(a^3)$ (multiplied by 3).
$\sqrt(a)=\sqrt(a^4)$ (multiplied by 4).
$\sqrt(a)*\sqrt(a)=\sqrt(a^3)*\sqrt(a^4)=\sqrt(a^3*a^4)=\sqrt(a^7)$.
Tasks for independent solution
1. Calculate: $\sqrt(32*243*1024)$.2. Calculate: $\sqrt(7\frac(58)(81))$.
3. Calculate:
a) $\sqrt(81)*\sqrt(72)$.
b) $\frac(\sqrt(1215))(\sqrt(5))$.
4. Simplify:
a) $\sqrt(\sqrt(a))$.
b) $\sqrt(\sqrt(a))$.
c) $\sqrt(\sqrt(a))$.
5. Perform actions: $\sqrt(a^2)*\sqrt(a^4)$.
To successfully use the operation of extracting the root in practice, you need to get acquainted with the properties of this operation.
All properties are formulated and proved only for non-negative values of variables contained under root signs.
Theorem 1. The nth root (n=2, 3, 4,...) of the product of two non-negative chipsets is equal to the product of the nth roots of these numbers:
Comment:
1.
Theorem 1 remains valid for the case when the radical expression is the product of more than two non-negative numbers.
Theorem 2.If a,
and n is a natural number greater than 1, then the equality
Brief(albeit inaccurate) formulation that is more convenient to use in practice: the root of the fraction is equal to the fraction of the roots.
Theorem 1 allows us to multiply m only roots of the same degree
, i.e. only roots with the same exponent.
Theorem 3. If ,k is a natural number and n is a natural number greater than 1, then the equality
In other words, to raise a root to a natural power, it is enough to raise the root expression to this power.
This is a consequence of Theorem 1. Indeed, for example, for k = 3 we get
Theorem 4. If ,k, n are natural numbers greater than 1, then the equality
In other words, to extract a root from a root, it is enough to multiply the exponents of the roots.
For example,
Be careful! We learned that four operations can be performed on roots: multiplication, division, exponentiation, and extracting the root (from the root). But what about the addition and subtraction of roots? No way.
For example, you can’t write instead of Indeed, But it’s obvious that
Theorem 5. If the indicators of the root and the root expression are multiplied or divided by the same natural number, then the value of the root will not change, i.e.
Examples of problem solving
Example 1 Calculate
Solution. Using the first property of the roots (Theorem 1), we get:
Example 2 Calculate
Solution. Convert the mixed number to an improper fraction.
We have Using the second property of the roots ( theorem 2
), we get:
Example 3 Calculate: 
Solution. Any formula in algebra, as you well know, is used not only "from left to right", but also "from right to left". So, the first property of the roots means that it can be represented as and, conversely, can be replaced by the expression. The same applies to the second property of roots. With this in mind, let's do the calculations.
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TOPIC: Power function. nth root
GOAL:
Repetition of the material covered during the game, conscious assimilation of these topics.
Education of responsibility, attention, memory training.
The development of ingenuity, resourcefulness. To promote the development of cognitive interest in mathematics.
ORGANIZING TIME
The bell rang. The children sat down in their places. The teacher asks questions to the students, and they answer the questions by raising their hands:
Can you please tell us what we studied in the last few lessons? ( the topic of this lesson the children name themselves)
What do you think is the purpose of our lesson today? ( The children try to formulate the purpose of the lesson themselves, the teacher only corrects it)
Welcome to the country"Mathematics "! To the country of logarithms, simple calculations, roots, erections and equations! Traveling across the countryMathematicians "2 commands are sent:" ROOT "," DEGREE ", the journey will take place under the motto (pre-written on the board ): “BOOK IS A BOOK, AND MOVE YOUR BRAIN” (V.V. Mayakovsky). Team members will be rewarded with "red cards" for correct answers.
1. Team building
Each student at the entrance to the office received a card on which the formula of the function is written (everyone has different ones). Each student determines which function he has, even or odd, if even - the “ROOT” command, odd - “DEGREE”.
Function options:f(x)=
,
f(x)=
f(x)= 
, f(x)= 
f(x)= 
f(x)= 
f(x)= 
f(x)= 
f(x)= 
f(x)= 
f(x)= 
, f(x)= 
f(x)= 
f(x)= 
f(x)=
f(x)=
f(x)=
f(x)=
2. The choice of the commander of each team
TASK: solve and defend your answer (the commander must be able to think quickly and be responsible for everything); for which values of the variable the expression makes sense ( expressions are written on the board in advance) :
|
Answer: -8≤ x Answer: -11≤ x
3. Warm up
For each correct answer - 1 card ( teams begin to score). The teacher reads the task, the students answer.
Arithmetic i sign
In the problem book you will find me in many lines.
Only "o" you insert into the word, knowing how,
And I am a geographical point. (+, pole)
I am a number less than ten
It's easy for you to find me.
But if you order the letter "I" to stand next to you,
I am everything - father, and you, and grandfather, and mother. (seven, family)
4. We continue the journey and on our way there is a huge wall on which the task is written (prepare a poster in the form of a wall in advance
): calculate: 
to overcome this wall, you need to solve this task, which team decides, that one will earn points.(0,7+0,3=1)
1) properties of a power function with n - even;
2) properties of a power function with n - odd.
6. The next test for us will be the "SHOW YOURSELF" contest. Conditions of the competition: each team member in turn goes to the board and solves any task of his choice, the first team to complete the tasks wins.
7. Teams prepare questions for each other. Receive points for the correct answer and for originality.
8. RESULT. AWARD. Each team prepares a final word, which reveals the following questions: what did today's lesson give each team and individual representatives, comments on the lesson and the teacher. Grading with comments (for what activity and why).
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