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 $\frac(0)(0)$.
Solution diagram standard examples 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 by 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 to solve standard tasks, to which we will now turn.
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 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 previous tasks. 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
Work helps you learn to extract Square root from any number without using a calculator and a table of squares and free the denominator of the fraction from irrationality.
Freeing yourself from the irrationality of the denominator of a fraction
The essence of the method is to multiply and divide fractions by an expression that will eliminate irrationality (square and cube roots) from the denominator and will make it simpler. After this, it is easier to reduce the fractions to 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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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 Latin word radix - root), fused with a superscript. In the old days, underlining an expression was used instead of the current bracketing, so there is just a modified ancient way records of something like. This designation was first used 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.
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 consider the basic rules of this transformation, and in the second - typical examples with detailed explanations.
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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 root signs there varying 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 to transform it, replacing it with 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's carry out necessary transformations and we obtain an identically equal fraction x · x + y x - y , freed 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 sequential conversion Fractions: Multiply both sides of a fraction by a number other than zero, and then convert the resulting denominator. Let's consider the main cases.
In the most simple case You can get by by transforming the denominator. For example, we can take a fraction with a denominator, equal to the root out 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
Previously in the article on transformations 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 it is this method. 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 area 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 previous paragraph we multiplied by conjugate expressions so that we could then use the difference of squares formula. 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 perform several transformations sequentially 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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Instructions
Before you get rid of irrationality V denominator, its type follows, and depending on this, continue the solution. And although any irrationality follows from simple presence, their various combinations and degrees presuppose different algorithms.
Presence below the line fractions root fractional power of the form m/n, with n>m This expression looks like this: a/√(b^m/n).
Get rid of this irrationality also by entering a multiplier, this time more complex: b^(n-m)/n, i.e. From the exponent of the root itself, you need the degree of the expression under its sign. Then in denominator only :a/(b^m/n) → a √(b^(n-m)/n)/b will remain. Example 2: 5/(4^3/5) → 5 √(4^2/5)/ 4 = 5 √(16^1/5)/4.
Sum of square rootsMultiply both components fractions by a similar difference. Then from the irrational addition of roots the denominator is transformed into / 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.
Sum/difference of cube rootsChoose the partial square of the difference as an additional factor if in denominator is the sum, and accordingly the incomplete square of the sum for the difference of the roots: a/(∛b ± ∛c) → a (∛b² ∓ ∛(b c) + ∛c²)/ ((∛b ± ∛c) ∛b² ∓ ∛(b c ) + ∛c²) →a (∛b² ∓ ∛(b c) + ∛c²)/(b ± c).Example 4: 7/(∛5 + ∛4) → 7 (∛25- ∛20 + ∛16) /9.
If the problem contains both square and , then divide the solution into two stages: successively derive the square root from the denominator, and then the cubic root. This is done using methods already known to you: in the first step you need to select the multiplier of the difference/sum of the roots, in the second - the incomplete square of the sum/difference.
Video on the topic
Sources:
- how to get rid of irrationality in fractions
Tip 2: How to get rid of irrationality in the denominator
Correct entry fractional number does not contain irrationality V denominator. Such a recording is easier to perceive by sight, so when irrationality V denominator It's wise to get rid of it. In this case, irrationality can become a numerator.
Instructions
To begin with, we can consider the simplest one - 1/sqrt(2). The square root of two is a number in . In this case, you need to multiply the numerator and denominator by its denominator. This will provide denominator. Indeed, sqrt(2)*sqrt(2) = sqrt(4) = 2. Multiplying two identical square roots by each other will ultimately give what is under each of the roots: in this case, two. As a result: 1/ sqrt(2) = (1*sqrt(2))/(sqrt(2)*sqrt(2)) = sqrt(2)/2. This algorithm also applies to fractions, in denominator whose root is multiplied by a rational 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.
You need to act in exactly the same way if denominator it is not the root that is found, but, say, the cubic or any other degree. Root in denominator you need to multiply by exactly the same root, and multiply the numerator by the same root. Then the root will go into the numerator.
In more cases, in denominator there is a sum of either an irrational number and or two irrational numbers. In the case of the sum (difference) of two square roots or a square root and rational number can be used well well-known formula(x+y)(x-y) = (x^2)-(y^2). It will help you get rid of 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 the conjugate to the expression in denominator.The effect of this 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.
If in denominator there is a sum (difference) in which the root is present to a greater extent, then the situation becomes nontrivial and getting rid of irrationality V denominator not always possible
Sources:
- get rid of the root in the denominator in 2019
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 by which it is divided 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, for example, the square root of two or pi. Usually when they talk about irrationality V denominator, the root is implied.
Instructions
Get rid of multiplying by the denominator. Thus it 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 the entire denominator is a root.
Multiply the numerator and denominator by the denominator the right number times, depending on the root. If the root is square, then once.
Multiply the numerator and denominator fractions to the denominator, that is, to √(x+2). The original example (56-y)/√(x+2) will turn into ((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 No irrationality.
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)).
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.
IN difficult cases repeat both of these options, applying as necessary. Please note that it is not always possible to get rid of irrationality V denominator.
Sources:
An algebraic fraction is an expression of the form A/B, where the letters A and B stand for any numeric or literal expressions. Often the numerator and denominator in algebraic fractions have a cumbersome appearance, but operations with such fractions should be performed according to the same rules as actions with ordinary ones, where the numerator and denominator are integers positive numbers.
Instructions
If given fractions, convert them (a fraction in which the numerator is greater 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.
If you need to convert a fraction to an improper fraction, then imagine it as numbers without a decimal point per one with as many zeros as there are numbers after the decimal point. For example, imagine the number 2.5 as 25/10 (if shortened, you get 5/2), and the number 3.61 as 361/100. It is often easier to operate with irregular ones than with mixed or decimal ones.
If you need to either subtract one fraction from another, and they have different denominators, bring the fractions to a common denominator. To do this, find the number that will be the least common multiple (LCM) of both denominators or several if there are more than two fractions. 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.
After the equal sign, swipe horizontal line and write this number (NOC) into the denominator. Add additional factors to each term - the number by which both the numerator and the denominator must be multiplied to obtain the LCM. Sequentially multiply the numerators by additional factors, maintaining the sign of addition or subtraction.
Calculate the result, shorten it if necessary, or select the entire part. For example, you need to add ⅓ and ¼. The LCM for both fractions is 12. Then the additional factor for the first fraction is 4, for the second - 3. Total: ⅓+¼=(1·4+1·3)/12=7/12.
If given for multiplication, multiply the numerators (this will be the numerator of the result) and the denominators (this will be the denominator of the result). In this case, there is no need to reduce them to a common denominator.
Factor the numerator and denominator as needed. For example, take the common factor out of brackets or use abbreviated multiplication formulas so that you can then, if necessary, reduce the numerator and denominator by gcd - the smallest common divisor.
note
Add numbers with numbers, letters of the same kind with letters of the same kind. For example, you cannot add 3a and 4b, which means that their sum or difference will remain in the numerator - 3a±4b.
Sources:
- Multiplying and dividing fractions
In everyday life we most often encounter integers: 1, 2, 3, 4, etc. (5 kg of potatoes), and fractional, non-integer numbers (5.4 kg of onions). Most of them are presented in form decimal fractions. But decimal submit to form fractions simple enough.
Instructions
For example, the number "0.12" is given. If not this fraction and imagine it as it is, then it will look like this: 12/100 (“twelve”). To get rid of a hundred in , you need to divide both the numerator and the denominator by the number that divides their numbers. This number is 4. Then, dividing the numerator and denominator, we get the number: 3/25.
If we consider a more everyday product, then it is often clear on the price tag 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 it were possible, this decimal fraction could be reduced further. And all using the same method: selecting a number that divides both the numerator and the denominator. This number is the largest common factor. The factor is “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.
At your request!
5. Solve the inequality:
6 . Simplify the expression:
17. f(x)=6x 2 +8x+5, F(-1)=3. Find F(-2).
Let's find C, knowing that F(-1) = 3.
3 = 2 ∙ (-1) 3 + 4 ∙ (-1) 2 + 5 ∙ (-1) + C;
3 = -2 + 4 – 5 + C;
Thus, the antiderivative F(x) = 2x 3 + 4x 2 + 5x + 6. Let’s find F(-2).
F(-2) = 2∙(-2) 3 +4∙(-2) 2 +5∙(-2)+6 = -16+16-10+6=-4.
20. Get rid of irrationality in the denominator
The solution is based on the basic property of a fraction, which allows you to multiply the numerator and denominator of a fraction by the same thing, without equal to zero number. To get rid of radical signs in the denominator of a fraction, they usually use abbreviated multiplication formulas. After all, if the difference of two radicals is multiplied by their sum, then we get the difference of the squares of the roots, i.e. you get an expression without radical signs.
21. Simplify the expression:
Let's solve this example in two ways. 1) Let's imagine the radical expression of the second factor in the form of the square of the sum of two expressions, i.e. in the form (a + b) 2 . This will allow us to extract the arithmetic square root.
2) Let's square the first factor and put it under the sign of the arithmetic square root of the second factor.
Decide in a way that is convenient for you!
22. Find (x 1 ∙y 1 +x 2 ∙y 2), where (x n; y n) are solutions to the system of equations:
Since the arithmetic square root can only be taken from non-negative number, That acceptable values variable at all numbers satisfying the inequality y≥0. Since the product in the first equation of the system is equal to negative number, then the following condition must be satisfied: x<0 . Let's express X from the first equation and substitute its value into the second equation. Let us solve the resulting equation for at, and then find the values X, corresponding to the previously obtained values at.
23. Solve the inequality: 7sin 2 x+cos 2 x>5sinx.
Since according to the main trigonometric identity: sin 2 x+cos 2 x=1, then presenting this inequality in the form 6sin 2 x+ sin 2 x +cos 2 x>5sinx and applying the main trigonometric identity, we get: 6sin 2 x+ 1>5sinx. Solving the inequality:
6sin 2 x-5sinx+1 >0. Let's make the replacement: sinx=y and get a quadratic inequality:
6y 2 -5y+1>0. Let us solve this inequality using the interval method by expanding left side by multipliers. To do this, we find the roots of the complete quadratic equation:
6y 2 -5y+1=0. Discriminant D=b 2 -4ac=5 2 -4∙6∙1=25-24=1. Then we get y 1 and y 2:
24. At the base of a straight prism lies regular triangle, whose area is Calculate the area of the lateral surface of the prism if its volume is 300 cm 3.
Let us be given the right one triangular prism ABCA 1 B 1 C 1, which is based on the correct Δ ABC, its area is known to us. Applying the area formula equilateral triangle, we will find our side triangle ABC. Since the volume of a straight prism is calculated by the formula V=S main. ∙ H, and we also know, then we can find H - the height of the prism. The lateral edge of the prism will be equal to the height of the prism: AA 1 =H. Knowing the side of the base and the length of the side edge of the prism, you can find the area of its lateral surface using the formula: S side. =P basic ∙ H.
25. There were 20 questions on the school quiz. For each correct answer, the participant was awarded 12 points, and for each incorrect answer, 10 points were deducted. How many correct answers did one of the participants give if he answered all the questions and scored 86 points?
Let the participant give x correct answers. Then he has (20) wrong answers. Knowing that for each correct answer he was awarded 12 points, and for each incorrect answer 10 points were deducted, and at the same time he scored 86 points, we will create the equation:
12x-10·(20's)=86;
12x-200+10x=86;
22x=286 ⇒ x=286:22 ⇒ x=13. The participant gave 13 correct answers.
I wish you to give 25 correct answers to the mathematics test at UNT!
24. In the right quadrangular pyramid height is 3, side rib 6. Find the radius of the sphere circumscribed about the pyramid.
Let a ball with center at point O 1 and radius MO 1 be described about regular pyramid MABCD with height MO=3 and side edge MA=6. It is required to find the radius of the ball MO 1. Consider ΔMAM 1, in which the side MM 1 is the diameter of the ball. Then ∠MAM 1 =90°. Let's find the hypotenuse MM 1 if the side MA and the projection of this side MO onto the hypotenuse are known. Remember? Height drawn from vertex right angle to the hypotenuse there is a mean proportional value between the projections of the legs onto the hypotenuse, and each leg is the average proportional value between the entire hypotenuse and the projection of this leg onto the hypotenuse. In this task, only the underlined part of the rule will be useful to us.
We write the equality: MA 2 =MO∙MM 1. We substitute our data: 6 2 =3∙MM 1. Hence MM 1 =36:3=12. We found the diameter of the ball, therefore, the radius of MO 1 =6.
25. Petya is older than Kolya, who is older than Misha, Masha is older than Kolya, and Dasha is younger than Petya, but older than Masha. Who is the third oldest?
Let's assume: older means more. Petya is older than Kolya, who is older than Misha Let's write it like this: Petya>Kolya>Misha. Dasha is younger than Petya, but older than Masha let's write it like this: Masha<Даша<Петя, что будет равнозначно записи: Петя>Dasha>Masha. Because Masha is older than Kolya, then we get: Petya>Dasha>Masha>Kolya. And finally: Petya>Dasha>Masha>Kolya>Misha. Thus, the third oldest is Masha.
I wish you successful preparation for the UNT!