When studying transformations of an irrational expression, a very important question is how to get rid of irrationality in the denominator of a fraction. The purpose of this article is to explain this action in specific examples tasks. In the first paragraph we will look at the basic rules of this transformation, and in the second - typical examples with detailed explanations.

The concept of liberation from irrationality in the denominator

Let's start by explaining what the meaning of such a transformation is. To do this, remember the following provisions.

We can talk about irrationality in the denominator of a fraction if there is a radical there, also known as the sign of the root. Numbers written using this sign are often irrational. Examples would be 1 2, - 2 x + 3, x + y x - 2 · x · y + 1, 11 7 - 5. Fractions with irrational denominators also include those that have signs of roots of various degrees (square, cubic, etc.), for example, 3 4 3, 1 x + x · y 4 + y. You should get rid of irrationality to simplify the expression and facilitate further calculations. Let's formulate the basic definition:

Definition 1

Free yourself from irrationality in the denominator of a fraction- means transforming it by replacing it with an identically equal fraction, the denominator of which does not contain roots or powers.

Such an action can be called liberation or getting rid of irrationality, but the meaning remains the same. So, the transition from 1 2 to 2 2, i.e. to a fraction with an equal value without a root sign in the denominator and will be the action we need. Let's give another example: we have a fraction x x - y. Let us carry out the necessary transformations and obtain an identically equal fraction x · x + y x - y , freeing ourselves from irrationality in the denominator.

After formulating the definition, we can proceed directly to studying the sequence of actions that need to be performed for such a transformation.

Basic steps to get rid of irrationality in the denominator of a fraction

To get rid of the roots, you need to carry out two successive transformations of the fraction: multiply both parts of the fraction by a number other than zero, and then transform the expression obtained in the denominator. Let's consider the main cases.

In the simplest case, you can get by by transforming the denominator. For example, we can take a fraction with a denominator equal to the root of 9. Having calculated 9, we write 3 in the denominator and thus get rid of irrationality.

However, much more often it is necessary to first multiply the numerator and denominator by a number that will then allow the denominator to be brought to the desired form (without roots). So, if we multiply 1 x + 1 by x + 1, we get the fraction x + 1 x + 1 x + 1 and can replace the expression in its denominator with x + 1. So we transformed 1 x + 1 into x + 1 x + 1, getting rid of the irrationality.

Sometimes the transformations you need to perform are quite specific. Let's look at a few illustrative examples.

How to convert an expression to the denominator of a fraction

As we said, the easiest way to do this is to convert the denominator.

Example 1

Condition: free the fraction 1 2 · 18 + 50 from the irrationality in the denominator.

Solution

First, let's open the brackets and get the expression 1 2 18 + 2 50. Using the basic properties of roots, we move on to the expression 1 2 18 + 2 50. We calculate the values ​​of both expressions under the roots and get 1 36 + 100. Here you can already extract the roots. As a result, we got the fraction 1 6 + 10, equal to 1 16. The transformation can be completed here.

Let's write down the progress of the entire solution without comment:

1 2 18 + 50 = 1 2 18 + 2 50 = 1 2 18 + 2 50 = 1 36 + 100 = 1 6 + 10 = 1 16

Answer: 1 2 18 + 50 = 1 16.

Example 2

Condition: given the fraction 7 - x (x + 1) 2. Get rid of irrationality in the denominator.

Solution

Earlier in the article devoted to transformations of irrational expressions using the properties of roots, we mentioned that for any A and even n we can replace the expression A n n with | A | over the entire range of permissible values ​​of variables. Therefore, in our case we can write it like this: 7 - x x + 1 2 = 7 - x x + 1. In this way we freed ourselves from irrationality in the denominator.

Answer: 7 - x x + 1 2 = 7 - x x + 1.

Getting rid of irrationality by multiplying by the root

If the denominator of a fraction contains an expression of the form A and the expression A itself does not have signs of roots, then we can free ourselves from irrationality by simply multiplying both sides of the original fraction by A. The possibility of this action is determined by the fact that A will not turn to 0 in the range of acceptable values. After multiplication, the denominator will contain an expression of the form A · A, which is easy to get rid of the roots: A · A = A 2 = A. Let's see how to correctly apply this method in practice.

Example 3

Condition: given fractions x 3 and - 1 x 2 + y - 4. Get rid of the irrationality in their denominators.

Solution

Let's multiply the first fraction by the second root of 3. We get the following:

x 3 = x 3 3 3 = x 3 3 2 = x 3 3

In the second case, we need to multiply by x 2 + y - 4 and transform the resulting expression in the denominator:

1 x 2 + y - 4 = - 1 x 2 + y - 4 x 2 + y - 4 x 2 + y - 4 = = - x 2 + y - 4 x 2 + y - 4 2 = - x 2 + y - 4 x 2 + y - 4

Answer: x 3 = x · 3 3 and - 1 x 2 + y - 4 = - x 2 + y - 4 x 2 + y - 4 .

If the denominator of the original fraction contains expressions of the form A n m or A m n (subject to natural m and n), we need to choose a factor such that the resulting expression can be converted to A n n k or A n k n (subject to natural k) . After this, it will be easy to get rid of irrationality. Let's look at this example.

Example 4

Condition: given fractions 7 6 3 5 and x x 2 + 1 4 15. Get rid of irrationality in denominators.

Solution

We need to take a natural number that can be divided by five, and it must be greater than three. In order for exponent 6 to become equal to 5, we need to multiply by 6 2 5. Therefore, we will have to multiply both parts of the original fraction by 6 2 5:

7 6 3 5 = 7 6 2 5 6 3 5 6 2 5 = 7 6 2 5 6 3 5 6 2 = 7 6 2 5 6 5 5 = 7 6 2 5 6 = 7 36 5 6

In the second case, we need a number greater than 15, which can be divided by 4 without a remainder. We take 16. To get such an exponent in the denominator, we need to take x 2 + 1 4 as a factor. Let us clarify that the value of this expression will not be 0 in any case. We calculate:

x x 2 + 1 4 15 = x x 2 + 1 4 x 2 + 1 4 15 x 2 + 1 4 = = x x 2 + 1 4 x 2 + 1 4 16 = x x 2 + 1 4 x 2 + 1 4 4 4 = x x 2 + 1 4 x 2 + 1 4

Answer: 7 6 3 5 = 7 · 36 5 6 and x x 2 + 1 4 15 = x · x 2 + 1 4 x 2 + 1 4 .

Getting rid of irrationality by multiplying by the conjugate expression

The following method is suitable for those cases when the denominator of the original fraction contains the expressions a + b, a - b, a + b, a - b, a + b, a - b. In such cases, we need to take the conjugate expression as a factor. Let us explain the meaning of this concept.

For the first expression a + b the conjugate will be a - b, for the second a - b – a + b. For a + b – a - b, for a - b – a + b, for a + b – a - b, and for a - b – a + b. In other words, a conjugate expression is an expression in which the opposite sign appears before the second term.

Let's look at what exactly this method is. Let's say we have a product of the form a - b · a + b. It can be replaced by the difference of squares a - b · a + b = a 2 - b 2, after which we move on to the expression a - b, devoid of radicals. Thus, we freed ourselves from irrationality in the denominator of the fraction by multiplying by the conjugate expression. Let's take a couple of illustrative examples.

Example 5

Condition: get rid of the irrationality in the expressions 3 7 - 3 and x - 5 - 2.

Solution

In the first case, we take the conjugate expression equal to 7 + 3. Now we multiply both parts of the original fraction by it:

3 7 - 3 = 3 7 + 3 7 - 3 7 + 3 = 3 7 + 3 7 2 - 3 2 = = 3 7 + 3 7 - 9 = 3 7 + 3 - 2 = - 3 7 + 3 2

In the second case, we need the expression - 5 + 2, which is the conjugate of the expression - 5 - 2. Multiply the numerator and denominator by it and get:

x - 5 - 2 = x · - 5 + 2 - 5 - 2 · - 5 + 2 = = x · - 5 + 2 - 5 2 - 2 2 = x · - 5 + 2 5 - 2 = x · 2 - 5 3

It is also possible to perform a transformation before multiplying: if we first remove the minus from the denominator, it will be more convenient to calculate:

x - 5 - 2 = - x 5 + 2 = - x 5 - 2 5 + 2 5 - 2 = = - x 5 - 2 5 2 - 2 2 = - x 5 - 2 5 - 2 = - x · 5 - 2 3 = = x · 2 - 5 3

Answer: 3 7 - 3 = - 3 7 + 3 2 and x - 5 - 2 = x 2 - 5 3.

It is important to pay attention to the fact that the expression obtained as a result of multiplication does not turn to 0 for any variables in the range of acceptable values ​​for this expression.

Example 6

Condition: given the fraction x x + 4. Transform it so that there are no irrational expressions in the denominator.

Solution

Let's start by finding the range of acceptable values ​​for the variable x. It is defined by the conditions x ≥ 0 and x + 4 ≠ 0. From them we can conclude that the desired region is a set x ≥ 0.

The conjugate of the denominator is x - 4 . When can we multiply by it? Only if x - 4 ≠ 0. In the range of acceptable values, this will be equivalent to the condition x≠16. As a result, we get the following:

x x + 4 = x x - 4 x + 4 x - 4 = = x x - 4 x 2 - 4 2 = x x - 4 x - 16

If x is equal to 16, then we get:

x x + 4 = 16 16 + 4 = 16 4 + 4 = 2

Therefore, x x + 4 = x · x - 4 x - 16 for all values ​​of x belonging to the range of acceptable values, with the exception of 16. At x = 16 we get x x + 4 = 2.

Answer: x x + 4 = x · x - 4 x - 16 , x ∈ [ 0 , 16) ∪ (16 , + ∞) 2 , x = 16 .

Converting fractions with irrationality in the denominator using sum and difference of cubes formulas

In the previous paragraph, we multiplied by conjugate expressions in order to then use the formula for the difference of squares. Sometimes, to get rid of irrationality in the denominator, it is useful to use other abbreviated multiplication formulas, for example, difference of cubes a 3 − b 3 = (a − b) (a 2 + a b + b 2). This formula is convenient to use if the denominator of the original fraction contains expressions with third-degree roots of the form A 3 - B 3, A 3 2 + A 3 · B 3 + B 3 2. etc. To apply it, we need to multiply the denominator of the fraction by the partial square of the sum A 3 2 + A 3 · B 3 + B 3 2 or the difference A 3 - B 3. The sum formula can be applied in the same way a 3 + b 3 = (a) (a 2 − a b + b 2).

Example 7

Condition: transform the fractions 1 7 3 - 2 3 and 3 4 - 2 · x 3 + x 2 3 so as to get rid of the irrationality in the denominator.

Solution

For the first fraction, we need to use the method of multiplying both parts by the partial square of the sum 7 3 and 2 3, since we can then convert using the difference of cubes formula:

1 7 3 - 2 3 = 1 7 3 2 + 7 3 2 3 + 2 3 2 7 3 - 2 3 7 3 2 + 7 3 2 3 + 2 3 2 = = 7 3 2 + 7 3 2 3 + 2 3 2 7 3 3 - 2 3 3 = 7 2 3 + 7 2 3 + 2 2 3 7 - 2 = = 49 3 + 14 3 + 4 3 5

In the second fraction we represent the denominator as 2 2 - 2 x 3 + x 3 2. This expression shows the incomplete square of the difference 2 and x 3, which means we can multiply both parts of the fraction by the sum 2 + x 3 and use the formula for the sum of cubes. To do this, the condition 2 + x 3 ≠ 0 must be met, equivalent to x 3 ≠ - 2 and x ≠ − 8:

3 4 - 2 x 3 + x 2 3 = 3 2 2 - 2 x 3 + x 3 2 = = 3 2 + x 3 2 2 - 2 x 3 + x 3 2 2 + x 3 = 6 + 3 x 3 2 3 + x 3 3 = = 6 + 3 x 3 8 + x

Let's substitute 8 into the fraction and find the value:

3 4 - 2 8 3 + 8 2 3 = 3 4 - 2 2 + 4 = 3 4

Let's summarize. For all x included in the range of values ​​of the original fraction (set R), with the exception of - 8, we get 3 4 - 2 x 3 + x 2 3 = 6 + 3 x 3 8 + x. If x = 8, then 3 4 - 2 x 3 + x 2 3 = 3 4.

Answer: 3 4 - 2 x 3 + x 2 3 = 6 + 3 x 3 8 + x, x ≠ 8 3 4, x = - 8.

Consistent application of different conversion methods

Often in practice there are more complex examples, when we cannot free ourselves from irrationality in the denominator using just one method. For them, you need to consistently perform several transformations or select non-standard solutions. Let's take one such problem.

Example N

Condition: convert 5 7 4 - 2 4 to get rid of the signs of the roots in the denominator.

Solution

Let's multiply both sides of the original fraction by the conjugate expression 7 4 + 2 4 with a non-zero value. We get the following:

5 7 4 - 2 4 = 5 7 4 + 2 4 7 4 - 2 4 7 4 + 2 4 = = 5 7 4 + 2 4 7 4 2 - 2 4 2 = 5 7 4 + 2 4 7 - 2

Now let's use the same method again:

5 7 4 + 2 4 7 - 2 = 5 7 4 + 2 4 7 + 2 7 - 2 7 + 2 = = 5 7 4 + 2 4 7 + 2 7 2 - 2 2 = 5 7 4 + 7 4 7 + 2 7 - 2 = = 5 7 4 + 2 4 7 + 2 5 = 7 4 + 2 4 7 + 2

Answer: 5 7 4 - 2 4 = 7 4 + 2 4 · 7 + 2.

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In this topic we will consider all three groups of limits with irrationality listed above. Let's start with limits containing uncertainty of the form $\frac(0)(0)$.

Uncertainty disclosure $\frac(0)(0)$.

The solution to standard examples of this type usually consists of two steps:

• We get rid of the irrationality that caused uncertainty by multiplying by the so-called “conjugate” expression;
• If necessary, factor the expression in the numerator or denominator (or both);
• We reduce the factors leading to uncertainty and calculate the desired value of the limit.

The term "conjugate expression" used above will be explained in detail in the examples. For now there is no reason to dwell on it in detail. In general, you can go the other way, without using the conjugate expression. Sometimes a well-chosen replacement can eliminate irrationality. Such examples are rare in standard tests, therefore, for the use of replacement, we will consider only one example No. 6 (see the second part of this topic).

We will need several formulas, which I will write down below:

\begin(equation) a^2-b^2=(a-b)\cdot(a+b) \end(equation) \begin(equation) a^3-b^3=(a-b)\cdot(a^2 +ab+b^2) \end(equation) \begin(equation) a^3+b^3=(a+b)\cdot(a^2-ab+b^2) \end(equation) \begin (equation) a^4-b^4=(a-b)\cdot(a^3+a^2 b+ab^2+b^3)\end(equation)

In addition, we assume that the reader knows the formulas for solving quadratic equations. If $x_1$ and $x_2$ are roots quadratic trinomial$ax^2+bx+c$, then it can be factorized using the following formula:

\begin(equation) ax^2+bx+c=a\cdot(x-x_1)\cdot(x-x_2) \end(equation)

Formulas (1)-(5) are quite sufficient for solving standard problems, which we will now move on to.

Example No. 1

Find $\lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)$.

Since $\lim_(x\to 3)(\sqrt(7-x)-2)=\sqrt(7-3)-2=\sqrt(4)-2=0$ and $\lim_(x\ to 3) (x-3)=3-3=0$, then in the given limit we have an uncertainty of the form $\frac(0)(0)$. The difference $\sqrt(7-x)-2$ prevents us from revealing this uncertainty. In order to get rid of such irrationalities, multiplication by the so-called “conjugate expression” is used. We will now look at how such multiplication works. Multiply $\sqrt(7-x)-2$ by $\sqrt(7-x)+2$:

$$(\sqrt(7-x)-2)(\sqrt(7-x)+2)$$

To open the brackets, apply , substituting $a=\sqrt(7-x)$, $b=2$ into the right side of the mentioned formula:

$$(\sqrt(7-x)-2)(\sqrt(7-x)+2)=(\sqrt(7-x))^2-2^2=7-x-4=3-x .$$

As you can see, if you multiply the numerator by $\sqrt(7-x)+2$, then the root (i.e., irrationality) in the numerator will disappear. This expression $\sqrt(7-x)+2$ will be conjugate to the expression $\sqrt(7-x)-2$. However, we cannot simply multiply the numerator by $\sqrt(7-x)+2$, because this will change the fraction $\frac(\sqrt(7-x)-2)(x-3)$, which is under the limit . You need to multiply both the numerator and denominator at the same time:

$$ \lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)= \left|\frac(0)(0)\right|=\lim_(x\to 3)\frac((\sqrt(7-x)-2)\cdot(\sqrt(7-x)+2))((x-3)\cdot(\sqrt(7-x)+2)) $$

Now remember that $(\sqrt(7-x)-2)(\sqrt(7-x)+2)=3-x$ and open the brackets. And after opening the parentheses and a small transformation $3-x=-(x-3)$, we reduce the fraction by $x-3$:

$$ \lim_(x\to 3)\frac((\sqrt(7-x)-2)\cdot(\sqrt(7-x)+2))((x-3)\cdot(\sqrt( 7-x)+2))= \lim_(x\to 3)\frac(3-x)((x-3)\cdot(\sqrt(7-x)+2))=\\ =\lim_ (x\to 3)\frac(-(x-3))((x-3)\cdot(\sqrt(7-x)+2))= \lim_(x\to 3)\frac(-1 )(\sqrt(7-x)+2) $$

The uncertainty $\frac(0)(0)$ has disappeared. Now you can easily get the answer of this example:

$$ \lim_(x\to 3)\frac(-1)(\sqrt(7-x)+2)=\frac(-1)(\sqrt(7-3)+2)=-\frac( 1)(\sqrt(4)+2)=-\frac(1)(4).$$

I note that the conjugate expression can change its structure, depending on what kind of irrationality it should remove. In examples No. 4 and No. 5 (see the second part of this topic) a different type of conjugate expression will be used.

Answer: $\lim_(x\to 3)\frac(\sqrt(7-x)-2)(x-3)=-\frac(1)(4)$.

Example No. 2

Find $\lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))$.

Since $\lim_(x\to 2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=\sqrt(2^2+5)-\sqrt(7\cdot 2 ^2-19)=3-3=0$ and $\lim_(x\to 2)(3x^2-5x-2)=3\cdot2^2-5\cdot 2-2=0$, then we we are dealing with uncertainty of the form $\frac(0)(0)$. Let's get rid of the irrationality in the denominator of this fraction. To do this, we add both the numerator and denominator of the fraction $\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))$ to the expression $\sqrt(x^ 2+5)+\sqrt(7x^2-19)$ conjugate to the denominator:

$$ \lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=\left|\frac(0 )(0)\right|= \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19))) ((\sqrt(x^2+5)-\sqrt(7x^2-19))(\sqrt(x^2+5)+\sqrt(7x^2-19))) $$

Again, as in example No. 1, you need to use parentheses to expand. Substituting $a=\sqrt(x^2+5)$, $b=\sqrt(7x^2-19)$ into the right side of the mentioned formula, we obtain the following expression for the denominator:

$$ \left(\sqrt(x^2+5)-\sqrt(7x^2-19)\right)\left(\sqrt(x^2+5)+\sqrt(7x^2-19)\ right)=\\ =\left(\sqrt(x^2+5)\right)^2-\left(\sqrt(7x^2-19)\right)^2=x^2+5-(7x ^2-19)=-6x^2+24=-6\cdot(x^2-4) $$

Let's return to our limit:

$$ \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)))((\sqrt(x ^2+5)-\sqrt(7x^2-19))(\sqrt(x^2+5)+\sqrt(7x^2-19)))= \lim_(x\to 2)\frac( (3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)))(-6\cdot(x^2-4))=\\ =-\ frac(1)(6)\cdot \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2-19)) )(x^2-4) $$

In example No. 1, almost immediately after multiplication by the conjugate expression, the fraction was reduced. Here, before the reduction, you will have to factorize the expressions $3x^2-5x-2$ and $x^2-4$, and only then proceed to the reduction. To factor the expression $3x^2-5x-2$ you need to use . First let's decide quadratic equation$3x^2-5x-2=0$:

$$ 3x^2-5x-2=0\\ \begin(aligned) & D=(-5)^2-4\cdot3\cdot(-2)=25+24=49;\\ & x_1=\ frac(-(-5)-\sqrt(49))(2\cdot3)=\frac(5-7)(6)=-\frac(2)(6)=-\frac(1)(3) ;\\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot3)=\frac(5+7)(6)=\frac(12)(6)=2. \end(aligned) $$

Substituting $x_1=-\frac(1)(3)$, $x_2=2$ into , we will have:

$$ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)(x-2)=3\cdot\left(x+\ frac(1)(3)\right)(x-2)=\left(3\cdot x+3\cdot\frac(1)(3)\right)(x-2) =(3x+1)( x-2). $$

Now it’s time to factorize the expression $x^2-4$. Let's use , substituting $a=x$, $b=2$ into it:

$$ x^2-4=x^2-2^2=(x-2)(x+2) $$

Let's use the results obtained. Since $x^2-4=(x-2)(x+2)$ and $3x^2-5x-2=(3x+1)(x-2)$, then:

$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x^2-5x-2)(\sqrt(x^2+5)+\sqrt(7x^2 -19)))(x^2-4) =-\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(x-2)(\sqrt(x ^2+5)+\sqrt(7x^2-19)))((x-2)(x+2)) $$

Reducing by the bracket $x-2$ we get:

$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(x-2)(\sqrt(x^2+5)+\sqrt(7x^ 2-19)))((x-2)(x+2)) =-\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(\sqrt( x^2+5)+\sqrt(7x^2-19)))(x+2). $$

All! The uncertainty has disappeared. One more step and we come to the answer:

$$ -\frac(1)(6)\cdot \lim_(x\to 2)\frac((3x+1)(\sqrt(x^2+5)+\sqrt(7x^2-19)) )(x+2)=\\ =-\frac(1)(6)\cdot\frac((3\cdot 2+1)(\sqrt(2^2+5)+\sqrt(7\cdot 2 ^2-19)))(2+2)= -\frac(1)(6)\cdot\frac(7(3+3))(4)=-\frac(7)(4). $$

Answer: $\lim_(x\to 2)\frac(3x^2-5x-2)(\sqrt(x^2+5)-\sqrt(7x^2-19))=-\frac(7)( 4)$.

In the following example, consider the case where irrationalities will be present in both the numerator and the denominator of the fraction.

Example No. 3

Find $\lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))$.

Since $\lim_(x\to 5)(\sqrt(x+4)-\sqrt(x^2-16))=\sqrt(9)-\sqrt(9)=0$ and $\lim_( x\to 5)(\sqrt(x^2-3x+6)-\sqrt(5x-9))=\sqrt(16)-\sqrt(16)=0$, then we have an uncertainty of the form $\frac (0)(0)$. Since in in this case Since the roots are present in both the denominator and the numerator, in order to get rid of uncertainty you will have to multiply by two brackets at once. First, to the expression $\sqrt(x+4)+\sqrt(x^2-16)$ conjugate to the numerator. And secondly, to the expression $\sqrt(x^2-3x+6)-\sqrt(5x-9)$ conjugate to the denominator.

$$ \lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))=\left|\frac(0)(0)\right|=\\ =\lim_(x\to 5)\frac((\sqrt(x+4)-\sqrt(x^2-16) )(\sqrt(x+4)+\sqrt(x^2-16))(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((\sqrt(x^2 -3x+6)-\sqrt(5x-9))(\sqrt(x^2-3x+6)+\sqrt(5x-9))(\sqrt(x+4)+\sqrt(x^2 -16))) $$ $$ -x^2+x+20=0;\\ \begin(aligned) & D=1^2-4\cdot(-1)\cdot 20=81;\\ & x_1=\frac(-1-\sqrt(81))(-2)=\frac(-10)(-2)=5;\\ & x_2=\frac(-1+\sqrt(81))( -2)=\frac(8)(-2)=-4. \end(aligned) \\ -x^2+x+20=-1\cdot(x-5)(x-(-4))=-(x-5)(x+4). $$

For the expression $x^2-8x+15$ we get:

$$ x^2-8x+15=0;\\ \begin(aligned) & D=(-8)^2-4\cdot 1\cdot 15=4;\\ & x_1=\frac(-(- 8)-\sqrt(4))(2)=\frac(6)(2)=3;\\ & x_2=\frac(-(-8)+\sqrt(4))(2)=\frac (10)(2)=5. \end(aligned)\\ x^2+8x+15=1\cdot(x-3)(x-5)=(x-3)(x-5). $$

Substituting the resulting expansions $-x^2+x+20=-(x-5)(x+4)$ and $x^2+8x+15=(x-3)(x-5)$ into the limit under consideration, will have:

$$ \lim_(x\to 5)\frac((-x^2+x+20)(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((x^2 -8x+15)(\sqrt(x+4)+\sqrt(x^2-16)))= \lim_(x\to 5)\frac(-(x-5)(x+4)(\ sqrt(x^2-3x+6)+\sqrt(5x-9)))((x-3)(x-5)(\sqrt(x+4)+\sqrt(x^2-16)) )=\\ =\lim_(x\to 5)\frac(-(x+4)(\sqrt(x^2-3x+6)+\sqrt(5x-9)))((x-3) (\sqrt(x+4)+\sqrt(x^2-16)))= \frac(-(5+4)(\sqrt(5^2-3\cdot 5+6)+\sqrt(5 \cdot 5-9)))((5-3)(\sqrt(5+4)+\sqrt(5^2-16)))=-6. $$

Answer: $\lim_(x\to 5)\frac(\sqrt(x+4)-\sqrt(x^2-16))(\sqrt(x^2-3x+6)-\sqrt(5x-9 ))=-6$.

In the next (second) part, we will consider a couple more examples in which the conjugate expression will have a different form than in the previous problems. The main thing to remember is that the purpose of using a conjugate expression is to get rid of the irrationality that causes uncertainty.

Tokarev Kirill

The work helps you learn how to extract the square root of any number without using a calculator and a table of squares and free the denominator of a fraction from irrationality.

Freeing yourself from the irrationality of the denominator of a fraction

The essence of the method is to multiply and divide a fraction by an expression that will eliminate irrationality (square and cube roots) from the denominator and make it simpler. After this, it is easier to reduce the fractions to a common denominator and finally simplify the original expression.

Extracting the square root with approximation to a given digit.

Suppose we need to extract the square root of the natural number 17358122, and it is known that the root can be extracted. To find the result, sometimes it is convenient to use the rule described in the work.

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Slide captions:

Radical. Freeing yourself from the irrationality of the denominator of a fraction. Extract the square root with a specified degree of accuracy. Student of class 9B of Municipal Educational Institution Secondary School No. 7, Salsk Kirill Tokarev

FUNDAMENTAL QUESTION: Is it possible to extract the square root of any number with a given degree of accuracy, without having a calculator and a table of squares?

GOALS AND OBJECTIVES: Consider cases of solving expressions with radicals that are not studied in school course mathematics, but necessary for the Unified State Exam.

HISTORY OF THE ROOT The root sign comes from the lowercase Latin letter r (initial in the Latin word radix - root), fused with a superscript. In the old days, underlining an expression was used instead of the current bracketing, so it is just a modified ancient way of writing something like. This notation was first used by the German mathematician Thomas Rudolf in 1525.

FREEDOM FROM IRRATIONALITY OF THE DENOMINATOR OF A FRACTION The essence of the method is to multiply and divide a fraction by an expression that will eliminate irrationality (square and cube roots) from the denominator and make it simpler. After this, it is easier to reduce the fractions to a common denominator and finally simplify the original expression. ALGORITHM FOR RELEASE FROM IRRATIONALITY IN THE DENOMINATOR OF A FRACTION: 1. Divide the denominator of the fraction into factors. 2. If the denominator has the form or contains a factor, then the numerator and denominator should be multiplied by. If the denominator is of the form or or contains a factor of this type, then the numerator and denominator of the fraction should be multiplied by or by, respectively. The numbers are called conjugates. 3. Convert the numerator and denominator of the fraction, if possible, then reduce the resulting fraction.

a) b) c) d) = - Liberation from irrationality in the denominator of the fraction.

EXTRACTING A SQUARE ROOT WITH APPROXIMATION TO A SPECIFIED DIGIT. 1) -1 100 96 400 281 11900 11296 24 4 281 1 2824 4 16 135 81 5481 4956 52522 49956 81 1 826 6 8326 6 2) Ancient Babylonian method: Example: Find. To solve the problem, this number is decomposed into the sum of two terms: 1700 = 1600 + 100 = 40 2 + 100, the first of which is a perfect square. Then we apply the formula. Algebraic way:

EXTRACTING A SQUARE ROOT WITH APPROXIMATION TO A SPECIFIED DIGIT. , 4 16 8 . 1 1 1 3 5 1 8 1 5 4 8 1 8 2 + 66 4 9 5 6 6 5 2 5 2 2 + 8 3 2 66 4 9 9 5 6 6 + 8 3 3 2 33 2 5 6 6 0 0 , 3

References 1. Collection of problems in mathematics for those entering universities, edited by M.I. Skanavi. V. K. Egerev, B. A. Kordemsky, V. V. Zaitsev, “ONICS 21st century”, 2003 2. Algebra and elementary functions. R. A. Kalnin, “Science”, 1973 3. Mathematics. Reference materials. V. A. Gusev, A. G. Mordkovich, publishing house “Prosveshcheniye”, 1990. 4. Schoolchildren about mathematics and mathematicians. Compiled by M.M. Liman, Enlightenment, 1981.

Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to understand in the most understandable way.
For example, you need to solve the simple equation x/b + c = d.

An equation of this type is called linear, because The denominator contains only numbers.

The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side cancels.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides by 5. We get:
x+20=45
x=45-20=25

Another example when the unknown is in the denominator:

Equations of this type are called fractional-rational or simply fractional.

We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic equation, which is solved in the usual way. You just need to consider the following points:

• the value of a variable that turns the denominator to 0 cannot be a root;
• You cannot divide or multiply an equation by the expression =0.

This is where the concept of the region of permissible values ​​(ADV) comes into force - these are the values ​​of the roots of the equation for which the equation makes sense.

Thus, when solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.

For example, you need to solve a fractional equation:

Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x – any value other than zero.

We get rid of the denominator by multiplying all terms of the equation by x

And we solve the usual equation

5x – 2x = 1
3x = 1
x = 1/3

Answer: x = 1/3

Let's solve a more complicated equation:

ODZ is also present here: x -2.

When solving this equation, we will not move everything to one side and bring the fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel out all the denominators at once.

To reduce the denominators, you need to multiply the left side by x+2, and the right side by 2. This means that both sides of the equation must be multiplied by 2(x+2):

This is the most common multiplication of fractions, which we have already discussed above.

Let's write the same equation, but slightly differently

The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain the usual linear equation:

x = 4 – 2 = 2, which corresponds to our ODZ

Answer: x = 2.

Solving equations with fractions not as difficult as it might seem. In this article we have shown this with examples. If you have any difficulties with how to solve equations with fractions, then unsubscribe in the comments.

There are several types irrationality fractions in the denominator. It is associated with the presence in it of an algebraic root of one or different degrees. In order to get rid of irrationality, it is necessary to perform certain mathematical operations depending on the situation.

Instructions

1. Before getting rid of irrationality fractions in the denominator, you should determine its type, and depending on this, continue the solution. Indeed, any irrationality follows from the simple presence of roots; their different combinations and degrees are assumed by different algorithms.

2. Square root of the denominator, expression of the form a/?bEnter an additional factor equal to?b. In order for the fraction not to change, it is necessary to multiply both the numerator and the denominator: a/?b ? (a ?b)/b.Example 1: 10/?3 ? (10?3)/3.

3. Presence below the line fractions a root of a fractional power of the form m/n, and n>mThis expression looks like this: a/?(b^m/n).

4. Get rid of similar irrationality also by entering a multiplier, this time more difficult: b^(n-m)/n, i.e. from the exponent of the root itself, it is necessary to subtract the degree of the expression under its sign. Then only the first power will remain in the denominator: a/(b^m/n) ? a ?(b^(n-m)/n)/b. Example 2: 5/(4^3/5) ? 5 ?(4^2/5)/4 = 5 ?(16^1/5)/4.

5. Sum square roots Multiply both components fractions by a similar difference. Then, from the irrational addition of roots, the denominator is transformed into the difference of expressions/numbers under the root sign: a/(?b + ?c) ? a (?b – ?c)/(b – c).Example 3: 9/(?13 + ?23) ? 9 (?13 – ?23)/(13 – 23) = 9 (?23 – ?13)/10.

6. Sum/difference of cube rootsChoose as an additional factor the incomplete square of the difference if the denominator contains a sum, and, accordingly, the incomplete square of the sum for the difference of the roots: a/(?b ± ?c) ? a (?b? ? ?(b c) + ?c?)/ ((?b ± ?c) ?b? ? ?(b c) + ?c?) ?a (?b? ? ?(b c) + ? c?)/(b ± c).Example 4: 7/(?5 + ?4) ? 7 (?25-?20 +?16)/9.

7. If the problem contains both a square and a cube root, then divide the solution into two stages: stepwise derive the square root from the denominator, and then the cubic root. This is done according to the methods already known to you: in the first action you need to choose the multiplier of the difference/sum of the roots, in the second - the incomplete square of the sum/difference.

Tip 2: How to get rid of irrationality in the denominator

Correct entry fractional number does not contain irrationality V denominator. Such a notation is easier to understand in appearance, therefore, when irrationality V denominator It's smart to get rid of it. In this case, irrationality can become a numerator.

Instructions

1. To begin with, let's look at a primitive example - 1/sqrt(2). Square root of 2 – irrational number V denominator.In this case, you need to multiply the numerator and denominator of the fraction by its denominator. This will provide a reasonable number in denominator. Indeed, sqrt(2)*sqrt(2) = sqrt(4) = 2. Multiplying 2 identical square roots by each other will result in what is under all of the roots: in this case, two. The result: 1/sqrt(2) = (1*sqrt(2))/(sqrt(2)*sqrt(2)) = sqrt(2)/2. This algorithm is also suitable for fractions, in denominator which the root is multiplied by a reasonable number. The numerator and denominator in this case must be multiplied by the root located in denominator.Example: 1/(2*sqrt(3)) = (1*sqrt(3))/(2*sqrt(3)*sqrt(3)) = sqrt(3)/(2*3) = sqrt( 3)/6.

2. Of course, something like this should be done if denominator It is not the square root that is found, but, say, the cubic root or any other degree. Root in denominator it is necessary to multiply by the same root, and the numerator is also multiplied by the same root. Then the root will go into the numerator.

3. In a more difficult case in denominator there is a sum or difference of an irrational and a reasonable number or 2 irrational numbers. In the case of the sum (difference) of 2 square roots or a square root and a reasonable number, you can use the famous formula (x+y)(x-y) = (x^2 )-(y^2). It will help you get rid of irrationality V denominator. If in denominator difference, then you need to multiply the numerator and denominator by the sum of the same numbers, if the sum - then by the difference. This multiplied sum or difference will be called conjugate to the expression in denominator.The result of this scheme is clearly visible in the example: 1/(sqrt(2)+1) = (sqrt(2)-1)/(sqrt(2)+1)(sqrt(2)-1) = (sqrt(2 )-1)/((sqrt(2)^2)-(1^2)) = (sqrt(2)-1)/(2-1) = sqrt(2)-1.

4. If in denominator there is a sum (difference) in which a root of a larger degree is present, then the situation becomes non-trivial and liberation from irrationality V denominator not invariably acceptable

Tip 3: How to free yourself from irrationality in the denominator of a fraction

A fraction consists of a numerator, located at the top of the line, and a denominator, the one it divides, located at the bottom. An irrational number is a number that cannot be represented in the form fractions with an integer in the numerator and a natural number in denominator. Such numbers are, say, the square root of 2 or pi. Traditionally, when talking about irrationality in denominator, the root is implied.

Instructions

1. Eliminate irrationality by multiplying by the denominator. This way the irrationality will be transferred to the numerator. When multiplying the numerator and denominator by the same number, the value fractions does not change. Use this option if each denominator is a root.

2. Multiply the numerator and denominator by the denominator the required number of times, depending on the root. If the root is square, then once.

3. Consider the example with square root. Take the fraction (56-y)/√(x+2). It has a numerator (56-y) and an irrational denominator √(x+2), which is the square root.

4. Multiply the numerator and denominator fractions to the denominator, that is, to √(x+2). The original example (56-y)/√(x+2) will become ((56-y)*√(x+2))/(√(x+2)*√(x+2)). The result will be ((56-y)*√(x+2))/(x+2). Now the root is in the numerator, and in denominator there is no irrationality.

5. Not invariably the denominator fractions each one is under the root. Get rid of irrationality by using the formula (x+y)*(x-y)=x²-y².

6. Consider the example with the fraction (56-y)/(√(x+2)-√y). Its irrational denominator contains the difference of 2 square roots. Complete the denominator to form (x+y)*(x-y).

7. Multiply the denominator by the sum of the roots. Multiply the numerator by the same to get the value fractions hasn't changed. The fraction will take the form ((56-y)*(√(x+2)+√y))/((√(x+2)-√y)*(√(x+2)+√y)).

8. Take advantage of the above property (x+y)*(x-y)=x²-y² and free the denominator from irrationality. The result will be ((56-y)*(√(x+2)+√y))/(x+2-y). Now the root is in the numerator, and the denominator has gotten rid of irrationality.

9. In difficult cases, repeat both of these options, using as necessary. Note that it is not always possible to get rid of irrationality in denominator .

An algebraic fraction is an expression of the form A/B, where the letters A and B stand for any number or letter expressions. Often the numerator and denominator in algebraic fractions have a massive appearance, but operations with such fractions should be done according to the same rules as actions with ordinary ones, where the numerator and denominator are positive integers.

Instructions

1. If given mixed fractions, convert them to irregular fractions (a fraction in which the numerator is larger than the denominator): multiply the denominator by the whole part and add the numerator. So the number 2 1/3 will turn into 7/3. To do this, multiply 3 by 2 and add one.

2. If you need to convert a decimal to an improper fraction, think of it as dividing a number without a decimal point by one with as many zeros as there are numbers after the decimal point. Let's say, imagine the number 2.5 as 25/10 (if you shorten it, you get 5/2), and the number 3.61 - as 361/100. Operating with improper fractions is often easier than with mixed or decimal fractions.

3. If fractions have identical denominators and you need to add them, then simply add the numerators; the denominators remain unchanged.

4. If you need to subtract fractions with identical denominators, subtract the numerator of the 2nd fraction from the numerator of the first fraction. The denominators also do not change.

5. If you need to add fractions or subtract one fraction from another, and they have different denominators, reduce the fractions to a common denominator. To do this, find a number that will be the least universal multiple (LCM) of both denominators or several if the fractions are larger than 2. LCM is a number that will be divided into the denominators of all given fractions. For example, for 2 and 5 this number is 10.

6. After the equal sign, draw a horizontal line and write this number (NOC) into the denominator. Add additional factors to the entire term - the number by which you need to multiply both the numerator and the denominator in order to get the LCM. Multiply the numerators step by step by additional factors, preserving the sign of addition or subtraction.

7. Calculate the total, reduce it if necessary, or select the entire part. For example, do you need to fold it? And?. The LCM for both fractions is 12. Then the additional factor for the first fraction is 4, for the 2nd fraction - 3. Total: ?+?=(1·4+1·3)/12=7/12.

8. If an example is given for multiplication, multiply the numerators together (this will be the numerator of the total) and the denominators (this will be the denominator of the total). In this case, there is no need to reduce them to a common denominator.

9. To divide a fraction by a fraction, you need to turn the second fraction upside down and multiply the fractions. That is, a/b: c/d = a/b · d/c.

10. Factor the numerator and denominator as needed. For example, move the universal factor out of the bracket or expand it according to abbreviated multiplication formulas, so that after this you can, if necessary, reduce the numerator and denominator by GCD - the minimum universal divisor.

Note!
Add numbers with numbers, letters of the same kind with letters of the same kind. Let's say it is impossible to add 3a and 4b, which means that their sum or difference will remain in the numerator - 3a±4b.

In everyday life, fake numbers are more common: 1, 2, 3, 4, etc. (5 kg of potatoes), and fractional, non-integer numbers (5.4 kg of onions). Many of them are presented in form decimals. But represent the decimal fraction in form fractions pretty easy.

Instructions

1. Let's say the number “0.12” is given. If you do not reduce this decimal fraction and present it as it is, then it will look like this: 12/100 (“twelve hundredths”). In order to get rid of a hundred in the denominator, you need to divide both the numerator and the denominator by a number that divides them into integers. This number is 4. Then, dividing the numerator and denominator, we get the number: 3/25.

2. If we look more at everyday life, we can often see on the price tag of products that its weight is, for example, 0.478 kg or so on. This number is also easy to imagine in form fractions:478/1000 = 239/500. This fraction is quite ugly, and if there was a probability, then this decimal fraction would be allowed to be reduced further. And all in the same way: selecting a number that divides both the numerator and the denominator. This number is called the greatest universal factor. The factor is named “largest” because it is much more convenient to immediately divide both the numerator and the denominator by 4 (as in the first example) than to divide it twice by 2.

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Decimal fraction– variety fractions, which has a “round” number in the denominator: 10, 100, 1000, etc., Say, fraction 5/10 has a decimal notation of 0.5. Based on this thesis, fraction can be represented as a decimal fractions .

Instructions

1. Possible, need to be represented as a decimal fraction 18/25. First, you need to make sure that one of the “round” numbers appears in the denominator: 100, 1000, etc. To do this, you need to multiply the denominator by 4. But you will need to multiply both the numerator and the denominator by 4.

2. Multiplying the numerator and denominator fractions 18/25 by 4, it turns out 72/100. This is recorded fraction in decimal form: 0.72.

When dividing 2 decimal fractions, when there is no calculator at hand, many experience some difficulties. There's really nothing difficult here. Decimal fractions are called such if their denominator has a number that is a multiple of 10. As usual, such numbers are written on one line and have a comma separating the fractional part from the whole. Apparently due to the presence of a fractional part, which also differs in the number of digits after the decimal point, it is not clear to many how to perform mathematical operations with such numbers without a calculator.

You will need

• sheet of paper, pencil

Instructions

1. It turns out that in order to divide one decimal fraction by another, you need to look at both numbers and determine which of them has more digits after the decimal point. We multiply both numbers by a number that is a multiple of 10, i.e. 10, 1000 or 100000, the number of zeros in which is equal to more digits later than the decimal point of one of our 2 initial numbers. Now both are decimal fractions turned into ordinary integers. Take a sheet of paper with a pencil and separate the two resulting numbers with a “corner”. We get the result.

2. Let's say we need to divide the number 7.456 by 0.43. The first number has more decimal places (3 decimal places), therefore we multiply both numbers not by 1000 and get two primitive integers: 7456 and 430. Now we divide 7456 by 430 with a “corner” and we get that if 7.456 is divided by 0.43 it will come out approximately 17.3.

3. There is another division method. Writing down decimals fractions in the form of primitive fractions with a numerator and denominator, for our case these are 7456/1000 and 43/100. Later, we write down the expression for dividing 2 primitive fractions: 7456*100/1000*43, after that we reduce the tens, we get: 7456/10*43 = 7456/430 In the final output we again get the division of 2 primitive numbers 7456 and 430, which can be produced with a “corner”.

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Helpful advice
Thus, the way to divide decimal fractions is to reduce them to whole numbers, with the support of multiplying each of them by the same number. Performing operations with integers, as usual, does not cause any difficulties for anyone.

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