When transforming a fractional algebraic expression whose denominator is written irrational expression, usually try to represent a fraction so that its denominator is rational. If A,B,C,D,... are some algebraic expressions, then you can specify rules with the help of which you can get rid of radical signs in the denominator of expressions of the form
In all these cases, liberation from irrationality is achieved by multiplying the numerator and denominator of the fraction by a factor chosen so that its product by the denominator of the fraction is rational.
1) To get rid of irrationality in the denominator of a fraction of the form . In multiply the numerator and denominator by
Example 1. .
2) In the case of fractions of the form . Multiply the numerator and denominator by an irrational factor
respectively, i.e. to the conjugate irrational expression.
The meaning of the last action is that in the denominator the product of the sum and the difference is transformed into a difference of squares, which will already be a rational expression.
Example 2. Free yourself from irrationality in the denominator of the expression:
Solution, a) Multiply the numerator and denominator of the fraction by the expression . We get (provided that)
3) In the case of expressions like
the denominator is considered as a sum (difference) and multiplied by partial square differences (sums) to obtain the sum (difference) of cubes ((20.11), (20.12)). The numerator is also multiplied by the same factor.
Example 3. Free yourself from irrationality in the denominator of expressions:
Solution, a) Considering the denominator of this fraction as the sum of the numbers and 1, multiply the numerator and denominator by the partial square of the difference of these numbers:
or finally:
In some cases, it is necessary to perform a transformation of the opposite nature: to free the fraction from irrationality in the numerator. It is carried out in exactly the same way.
Example 4. Free yourself from irrationality in the numerator of a fraction.
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 $\frac(\sqrt(7-x)-2)(x-3)$ 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 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 next example, consider the case where irrationalities will be present in both the numerator and 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 this case 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, we 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.
Expressions, expression conversion
How to free yourself from irrationality in the denominator? Methods, examples, solutions
In 8th grade, during algebra lessons, within the framework of the topic of transformation of irrational expressions, the conversation turns to liberation from irrationality in the denominator of a fraction. In this article we will analyze what kind of transformation this is, consider what actions allow you to free yourself from irrationality in the denominator of a fraction, and provide solutions to typical examples with detailed explanations.
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What does it mean to free yourself from irrationality in the denominator of a fraction?
First you need to understand what irrationality is in the denominator and what it means to free yourself from irrationality in the denominator of a fraction. Information from school textbooks will help us with this. The following points deserve attention.
When the notation of a fraction contains a root sign (radical) in the denominator, then it is said that the denominator contains irrationality. This is probably due to the fact that numbers written using root signs are often . As an example, we give the fractions , , 
, obviously, the denominators of each of them contain the sign of the root, and therefore irrationality. In high school, it is inevitable to encounter fractions, the irrationality in the denominators of which is introduced not only by signs square roots, but also signs of cube roots, fourth roots, etc. Here are examples of such fractions: , 
.
Considering the information provided and the meaning of the word “free”, the following definition is very natural:
Definition.
Liberation from irrationality in the denominator of a fraction is a transformation in which a fraction with an irrationality in the denominator is replaced by an identically equal fraction that does not contain root signs in the denominator.
You can often hear people say not to free themselves, but to get rid of the irrationality in the denominator of the fraction. The meaning does not change.
For example, if we move from a fraction to a fraction whose value is equal to the value of the original fraction and the denominator of which does not contain the root sign, then we can state that we have freed ourselves from irrationality in the denominator of the fraction. Another example: replacing a fraction with an identical fraction 
there is a liberation from irrationality in the denominator of the fraction.
So, the initial information has been received. It remains to find out what needs to be done to free ourselves from irrationality in the denominator of the fraction.
Ways to free yourself from irrationality, examples
Usually, to get rid of irrationality, two are used in the denominator of a fraction. fraction conversions: Multiplying the numerator and denominator by a non-zero number or expression and transforming the expression in the denominator. Below we'll look at how these fraction conversions are used in basic ways to remove irrationality from the denominator of a fraction. Let's touch on the following cases.
In the simplest cases, it is enough to transform the expression into the denominator. An example is a fraction whose denominator is the root of nine. In this case, replacing it with the value 3 frees the denominator from irrationality.
In more difficult cases you have to first multiply the numerator and denominator of the fraction by some non-zero number or expression, which subsequently allows you to convert the denominator of the fraction to a form that does not contain radical signs. For example, after multiplying the numerator and denominator of a fraction by , the fraction takes the form 
, and then the expression in the denominator can be replaced by an expression without signs of the roots x+1. Thus, after being freed from irrationality in the denominator, the fraction takes the form .
If we talk about the general case, then in order to get rid of irrationality in the denominator of a fraction, one has to resort to various permissible transformations, sometimes quite specific ones.
And now in detail.
Converting an Expression to the Denominator of a Fraction
As already noted, one way to get rid of irrationality in the denominator of a fraction is to transform the denominator. Let's look at the solutions to the examples.
Example.
Get rid of irrationality in the denominator of a fraction 
.
Solution.
Opening the parentheses in the denominator, we arrive at the expression 
. Then they allow you to move on to fractions 
. Having calculated the values under the signs of the roots, we have 
. Obviously, in the resulting expression it is possible, which gives a fraction that is equal to 1/16. This is how we got rid of irrationality in the denominator.
Usually the solution is written briefly without explanation, since the actions performed are quite simple: 
Answer:
.
Example.
Solution.
When we talked about transforming irrational expressions using the properties of roots, we noted that for any expression A with even n (in our case n=2) the expression can be replaced by the expression |A| on the entire ODZ of variables for the original expression. Therefore, you can perform the following transformation of a given fraction: 
, which frees us from irrationality in the denominator.
Answer:
.
Multiplying the numerator and denominator by the root
When the expression in the denominator of a fraction has the form , where the expression A does not contain signs of the roots, then multiplying the numerator and denominator by allows you to get rid of irrationality in the denominator. This action is possible because it does not vanish on the variable variables for the original expression. In this case, the denominator produces an expression that can be easily converted to a form without radical signs: 
. Let us demonstrate the application of this approach with examples.
Example.
Free yourself from irrationality in the denominator of the fraction: a) , b) .
Solution.
a) Multiplying the numerator and denominator of the fraction by square root out of three, we get 
.
b) To get rid of the square root sign in the denominator, multiply the numerator and denominator of the fraction by , and then carry out transformations in the denominator: 
Answer:
a) , b) 
.
In the case when the denominator contains factors or , where m and n are some natural numbers, the numerator and denominator must be multiplied by such a factor so that after this the expression in the denominator can be converted to the form or , where k is some natural number, respectively. Then it’s easy to move on to a fraction without irrationality in the denominator. Let us demonstrate the application of the described method of getting rid of irrationality in the denominator using examples.
Example.
Free yourself from irrationality in the denominator of the fraction: a) , b) .
Solution.
a) The nearest natural number greater than 3 and divisible by 5 is 5. In order for the exponent of six to become equal to five, the expression in the denominator must be multiplied by. Consequently, liberation from irrationality in the denominator of a fraction will be facilitated by the expression by which the numerator and denominator must be multiplied: 
b) Obviously, the nearest natural number that exceeds 15 and is divisible by 4 without a remainder is 16. To get the exponent in the denominator to become equal to 16, you need to multiply the expression there by. Thus, multiplying the numerator and denominator of the original fraction by (note, the value of this expression is not equal to zero for any real x) will get rid of the irrationality in the denominator: 
Answer:
A) 
, b) 
.
Multiplying by its conjugate
The following method of getting rid of irrationality in the denominator of a fraction covers cases when the denominator contains expressions of the form , , , , or . In these cases, in order to free yourself from irrationality in the denominator of the fraction, you need to multiply the numerator and denominator of the fraction by the so-called conjugate expression.
It remains to find out which expressions are conjugate to the above. For an expression, the conjugate expression is , and for an expression, the conjugate expression is . Similarly, for an expression the conjugate is , and for an expression the conjugate is . And for an expression the conjugate is , and for an expression the conjugate is . So, the expression conjugate to this expression differs from it by the sign in front of the second term.
Let's see what multiplying an expression by its conjugate results in. For example, consider the work 
. It can be replaced by the difference of squares, that is, , from where we can then move on to the expression a−b, which does not contain signs of the roots.
Now it becomes clear how multiplying the numerator and denominator of a fraction by the expression conjugate to the denominator allows you to free yourself from irrationality in the denominator of the fraction. Let's look at solutions to typical examples.
Example.
Imagine the expression as a fraction whose denominator does not contain a radical: a) , b) .
Solution.
a) The expression conjugate to the denominator is . Let's multiply the numerator and denominator by it, which will allow us to free ourselves from irrationality in the denominator of the fraction: 
b) The conjugate of the expression is . Multiplying the numerator and denominator by it, we get 
It was possible to first remove the minus sign from the denominator, and only after that multiply the numerator and denominator by the expression conjugate to the denominator: 
Answer:
A) 
, b) 
.
Please note: when multiplying the numerator and denominator of a fraction by an expression with variables conjugate to the denominator, care must be taken that it does not vanish for any set of values of the variables from the ODZ for the original expression.
Example.
Free yourself from irrationality in the denominator of a fraction.
Solution.
First, let's find the range of permissible values (APV) of the variable x. It is determined by the conditions x≥0 and , from which we conclude that the ODZ is the set x≥0.
The expression conjugate to the denominator is . We can multiply the numerator and denominator of the fraction by it, provided that , which on the ODZ is equivalent to the condition x≠16. In this case we have 
And at x=16 we have 
.
Thus, for all values of the variable x from the ODZ, except x=16, 
, and for x=16 we have .
Answer:
Using the sum of cubes and difference of cubes formulas
From the previous paragraph, we learned that multiplying the numerator and denominator of a fraction by the expression conjugate to the denominator is carried out in order to subsequently apply the difference of squares formula and thereby free ourselves from irrationality in the denominator. In some cases, other abbreviated multiplication formulas are useful for getting rid of irrationality in the denominator. For example, the formula for difference of cubes a 3 −b 3 =(a−b)·(a 2 +a·b+b 2) allows you to get rid of irrationality when the denominator of a fraction contains expressions with cubic roots of the form or 
, where A and B are some numbers or expressions. To do this, the numerator and denominator of the fraction are multiplied by the partial square of the sum or by the difference, respectively. The formula for sum of cubes is used in the same way. a 3 +b 3 =(a+b)·(a 2 −a·b+b 2).
Example.
Free yourself from irrationality in the denominator of the fraction: a) , b) 
.
Solution.
a) It’s easy to guess that in this case, multiplying the numerator and denominator by the incomplete square of the sum of the numbers and allows you to free yourself from irrationality in the denominator, since in the future this will allow you to transform the expression in the denominator using the difference of cubes formula: 
b) Expression in the denominator of the fraction can be represented in the form 
, from which it is clearly seen that this is the incomplete square of the difference between the numbers 2 and . Thus, if the numerator and denominator of a fraction are multiplied by the sum, then the denominator can be converted using the sum of cubes formula, which will free us from irrationality in the denominator of the fraction. This can be done under the condition that is equivalent to the condition further x≠−8: 
And when substituting x=−8 into the original fraction we have 
.
Thus, for all x from the ODZ for the original fraction (in this case this is the set R), except x=−8, we have 
, and for x=8 we have 
.
Answer:
Using different methods
In more complex examples, it is usually not possible to free yourself from irrationality in the denominator in one action, but you have to consistently apply method after method, including those discussed above. Sometimes some non-standard solutions may be required. Enough interesting tasks on the topic under discussion can be found in the textbook authored by Yu. N. Kolyagin. References.
- Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
- Algebra and started mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.
Danny Peric Campana
One more interesting book for schoolchildren interested in, unfortunately, not translated into Russian, this is the book “Daniel’s Mathematical Adventures” (Las Aventuras Matemáticas de Daniel) by the Chilean mathematics teacher Danny Perich Campana, a very extraordinary and interesting person. He not only teaches children, but also writes songs and posts various educational materials on mathematics on the Internet. They can be found on YouTube and on the website http://www.sectormatematica.cl/ (of course, all materials are in Spanish).
Here I am posting one chapter from Danni Peric’s book. I found it quite interesting and useful for schoolchildren. To make it clear what we are talking about, I will say that Daniel and Camila work at the school, they are teachers.
The secret of getting rid of irrationality
“Camila, I’m having a lot of problems now when I’m trying to explain why what we’re going through in class is used,” said Daniel.
- I don’t really understand what you’re talking about.
— I’m talking about what is in everyone school textbooks and even university-level books. I still have doubts: why do we need to get rid of irrationality in the denominator? And I hate telling people what I haven’t understood for so long,” Daniel complained.
“I also don’t know where this comes from and why it is needed, but there must be some logical explanation for this.
— I once read in a scientific journal that getting rid of irrationality in the denominator allows you to obtain the result with greater accuracy, but I have never seen this again and am not sure that this is true.
- Why don’t we check it? - Camila asked.
“You’re right,” Daniel agreed. — Instead of complaining, you should try to draw your own conclusions. Then help me...
- Of course, now I’m interested in this myself.
“We have to take some expressions and get rid of the irrationality in the denominator, then replace the root with its value and find the result of the expression before and after getting rid of the irrationality in the denominator and see if anything changes.”
“Of course,” Camila agreed. - Let's do that.
“Take, for example, the expression,” said Daniel and took a piece of paper to write down what was happening. - Multiply the numerator and denominator by and get .
“It will be correct and can help us draw conclusions if we consider other irrational expressions equal to this one,” Camila suggested.
“I agree,” said Daniel, “I will divide the numerator and denominator by , and you multiply them by .”
- I did it. And you?
“I have,” Daniel answered. - Now let’s calculate the original expression and the resulting ones, replacing it with its value with all the decimal places that the calculator gives. We get:
“I don’t see anything special,” said Camila. “I was expecting some kind of difference that would justify getting rid of irrationality.”
“As I already told you, I once read about this in connection with the approach. What do you say if we replace it with a less precise number, such as ?
- Let's try and see what happens.
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 take the square root of natural number 17358122, and it is known that the root is extracted. To find the result, sometimes it is convenient to use the rule described in the work.
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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 given number decomposes into the sum of two terms: 1700 = 1600 + 100 = 40 2 + 100, the first of which is 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.
