Root of the nth degree examples for self-decision. Square root

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

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: $\sqrt(7\frac(19)(32))$.
Decision. 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) $\frac(\sqrt(256))(\sqrt(4))$.
Decision:
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 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: $\sqrt(a)*\sqrt(a)$.
Decision.
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

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. Root nth degree(n=2, 3, 4,...) from the product of two non-negative chips is equal to the product roots of the nth degrees from 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

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 multiply or divide the indicators of the root and the root expression by the same natural number, then the value of the root will not change, i.e.

Examples of problem solving

Example 2 Calculate
Decision. Convert the mixed number to an improper fraction.
We have Using the second property of the roots ( theorem 2 ), we get:

Example 3 Calculate:

Decision. 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.

We need to get acquainted with the properties of this operation, which we will do in this section.

All properties are formulated and proved only for non-negative values ​​of variables contained under root signs.

Proof. Let us introduce the following notation: We need to prove that for non-negative numbers x, y, z the equality x-yz holds.
As
So, But if the degrees of two non-negative numbers are equal and the exponents are equal, then the bases are also equal degrees; hence, from the equality x n \u003d (yz) p it follows that x-yz, and this was required to be proved.

We give a brief record of the proof of the theorem.

Notes:

1. Theorem 1 remains valid for the case when the radical expression is the product of more than two non-negative numbers.
2. Theorem 1 can be formulated using the "if...then" construction (as is customary for theorems in mathematics). Let us give the corresponding formulation: if a and b are non-negative numbers, then equality is true. We will formulate the following theorem in exactly this way.

A short (albeit inaccurate) formulation that is more convenient to use in practice: the root of fractions is equal to the fraction of the roots.

Proof. We give a brief record of the proof of Theorem 2, and you try to make the appropriate comments similar to those given in the proof of Theorem 1.

You, of course, noticed that the proven two properties n-th roots degrees are a generalization of the properties of square roots known to you from the 8th grade algebra course. And if other properties of roots nth degree was not, then everything would be simple (and not very interesting). In fact, there are several other interesting and important properties that we will discuss in this paragraph. But first, let's look at some examples of using Theorems 1 and 2.

Example 1 Calculate
Decision. Using the first property of the roots (Theorem 1), we get:

Remark 3. You can, of course, solve this example differently, especially if you have a microcalculator at hand: multiply the numbers 125, 64 and 27, and then extract the cube root from the resulting product. But, you see, the proposed solution is "smarter".
Example 2 Calculate
Decision. Convert the mixed number to an improper fraction.
We have Using the second property of the roots (Theorem 2), we get:

Example 3 Calculate:
Decision. 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:

Example 4 Run actions:
Decision, a) We have:
b) Theorem 1 allows us to multiply only roots of the same degree, i.e. only roots with the same exponent. Here it is also proposed to multiply the root of the 2nd degree from the number a by the root of the 3rd degree of the same number. How to do this, we do not yet know. We will return to this problem later.
Let us continue studying the properties of radicals.

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

In other words, to extract a root from a root, it is enough to multiply the exponents of the roots.
For example,
Proof. As in Theorem 2, we give a brief record of the proof, and you can try to make the appropriate comments on your own, similar to those given in the proof of Theorem 1.

Remark 4. Let's take a breath. What have we learned from proven theorems? 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. We talked about this back in the 8th grade about the operation of extracting a square root.

For example, you can’t write instead of Indeed, But it’s obvious that Be careful!
Perhaps the most interesting property of roots is the one that will be discussed in the next theorem. Considering the special significance of this property, we take the liberty of violating the certain style of formulations and proofs developed in this section in order to make the formulation of Theorem 5 a little "softer" and its proof more understandable.

For example:

(indicators of the root and root expression divided by 4);

(indicators of the root and root expression divided by 3);

(the indicators of the root and radical expression were multiplied by 2).

Proof. Let us denote the left side of the equality to be proved by the letter Then, by the definition of the root, the equality

Denote the right side of the identity being proved by the letter y:

Then, by the definition of a root, the equality

Let us raise both sides of the last equality to the same power p; we get:

So (see equalities (1) and (2)),

Comparing these two equalities, we come to the conclusion that x np = y np, and hence x = y, which was to be proved.
The proved theorem will allow us to solve the problem that we encountered above when solving example 5, where it was required to perform the multiplication of roots with different exponents:

This is how it is usually argued in such cases.
1) According to Theorem 5 in the expression, it is possible to multiply both the root index (i.e. the number 2) and the index of the root expression (i.e. the number 1) by the same natural number. Using this, we multiply both indicators by 3; we get:
2) According to Theorem 5, in the expression it is possible to multiply both the root index (i.e. the number 3) and the index of the root expression (i.e. the number 1) by the same natural number. Using this, we multiply both indicators by 2; we get:

3) Since we got the roots of the same 6th degree, we can multiply them:

Remark 5. Have you forgotten that all the properties of the roots that we discussed in this paragraph are considered by us only for the case when the variables take only non-negative values? Why did you have to make such a restriction? Because nth root degree from a negative number does not always make sense - it is defined only for odd values ​​of n. For such values ​​of the root exponent, the considered properties of the roots are also true in the case of negative radical expressions.

A.G. Mordkovich Algebra Grade 10

SUBJECT: Power function. Root nth degrees

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.

Compare:

1)

2)

3)

Solve the equation:

4)

6)

Calculate:

7)

8)

9)

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 was useful today's lesson for each team and individual representatives, comments on the lesson and the teacher. Grading with comments (for what activity and why).