Simplification Of Algebraic Expressions
Sumita has a 20 meter long red ribbon. Santanu and I have made some cards. We have decided that we will cover the four sides of the cards by the red ribbon.
If Sumita has a 4x2meter long red ribbon and Santanu and I took ax meter and 2xb meter respectively from Sumita’s ribbon, then let’s calculate how much portion we took from Sumita’s ribbon.
1 took \(\frac{2 x b}{4 x^2}\) Part = \(\frac{b}{2 x}\)2 Par
I have seen that \(\frac{2 x b}{4 x^2}\) and \(\frac{b}{2 x}\) are same. What are these called?
Santanu took \(\frac{a x}{4 x^2}\) part = \(\frac{a}{4 x}\) part.
We took, \(\frac{b}{2 x}\) part + \(\frac{a}{4 x}\) part.
= \(\left(\frac{b}{2 x}+\frac{a}{4 x}\right)\) part = \(\left(\frac{2 b+a}{4 x}\right)\) part of the total ribbon. Let’s simplify the algebraic expression.
Question 1. \(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)
Solution:
Given
\(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)First we see the L.C.M. of (b-c) (c-a), (c-a) (a-b) and (a-b) (b-c) (b-c) (c-a), (c-a) (a-b) and (a-b) (b-c) = (a-b) (b-c) (c-a)
= \(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)
= \(\frac{a-b}{(a-b)(b-c)(c-a)}+\frac{b-c}{(a-b)(b-c)(c-a)}+\frac{c-a}{(a-b)(b-c)(c-a)}\)
(a-b)(b-c)(c-a)÷(b-c)(c-a)=(a-b)
(a-b)(b-c)(c-a)÷(c-a)(a-b)=(b-c)
(a-b)(b-c)(c-a)÷(a-b)(b-c)=(c-a)
= \(\frac{a-b+b-c+b-a}{(a-b)(b-c)(c-a)}=\frac{0}{(a-b)(b-c)(c-a)}=0\)
\(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\) = 0
Read and Learn More WBBSE Solutions For Class 8 Maths
Question 2. Let the length of the ribbon is 8a²x2meter. I took 2ax meter ribbon.
Solution:
Given
The length of the ribbon is 8a²x2meter. I took 2ax meter ribbon
I took \(\frac{2 a x}{8 a^2 x^2}\) part = \(\frac{1}{4 a x}\) part
∴ The reduced from of \(\frac{2 a x}{8 a^2 x^2}\) is \(\frac{1}{4 a x}\)
Question 3. Let’s express \(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\) in reduced form-
Solution:
Given
\(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\)= \(\frac{a+1}{a+2} \times \frac{a^2-a-2}{a^2+\bar{a}}=\frac{a+1}{a+2} x\)
= \(=\frac{(a-2) \times(a+1)}{a(a+1)}\)
= \(\frac{(a+1)(a-2)}{a(a+2)}\)
\(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\) = \(\frac{(a+1)(a-2)}{a(a+2)}\)
Algebraic Expressions Exercise
Question 1. Let’s express \(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\) in reduced form.
Solution:
Given
\(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\)= \(\frac{a(a+b)}{a-b} \times \frac{a-b}{b(a+b)} \times \frac{a}{b}\)
= \(\frac{a^2(a+b)(a-b)}{b^2(a-b)(a+b)}=\frac{a^2}{b^2}\)
\(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\) = \(\frac{a^2(a+b)(a-b)}{b^2(a-b)(a+b)}=\frac{a^2}{b^2}\)
Question 2. Let’s express \(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\) in reduced form.
Solution:
Given
\(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\)= \(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}=\frac{x^2+x-2}{x^2-2 x-8} \times \frac{x^2-3 x-4}{x^2-x-6}\)
= \(\frac{x^2+2 x-x-2}{x^2-4 x+2 x-8} \times \frac{x^2-4 x+x-4}{x^2-3 x+2 x-6}\)
= \( \frac{x(x+2)-1(x+2)}{x(x-4)+2(x-4)} \times \frac{x(x-4)+1(x-4)}{x(x-3)+2(x-3)}\)
= \(\frac{(x+2)(x-1)}{(x-4)(x+2)} \times \frac{(x-4)(x+1)}{(x-3)(x+2)}\)
= \(\frac{(x+2)(x-1)(x-4)(x+1)}{(x-4)(x+2)(x-3)(x+2)}\)
[In numerator and in denominator [x+2×x-4] is the common factor, so dividing numerator and denominator by (x+2) (x-4) we get the reduced form.]
= \(\frac{(x+2)(x-1)(x-4)(x+1)}{(x-4)(x+2)(x-3)(x+2)}\)
= \(\frac{(x-1)(x+1)}{(x-3)(x+2)}\)
\(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\) = \(\frac{(x-1)(x+1)}{(x-3)(x+2)}\)
Question 3. Let’s express \(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\) in reduced form and let’s write the common factor of the numerator and the denominator.
Solution:
Given
\(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\)= \(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\)
= \(\frac{p^2-q^2}{x-y} \div \frac{x^2-y^2}{p+q}\)
= \(\frac{(p+q)(p-q)}{(x-y)} \times \frac{(x-y)(x+y)}{p+q}\)
\(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\) = (p-q)(x+y)
The common factor of the numerator and the denominator is (p-q)(x+y)
Algebraic Expressions Exercise
Let’s express the algebraic expressions given below in reduced form:
1. \(\frac{a^2 \times c^2}{c^2 \times d^2} \div \frac{b c}{a d}\)
Solution:
Given
= \(\frac{a^2 \times c^2}{c^2 \times d^2}\)
= \(\frac{a^3}{b c d}\)
\(\frac{a^2 \times c^2}{c^2 \times d^2} \div \frac{b c}{a d}\) = \(\frac{a^3}{b c d}\)
2. \(\frac{x^2 y-x y^2}{x^2-x y}\)
Solution:
Given
= \(\frac{x^2 y-x y^2}{x+y}\)
= \(\frac{x y(x-y)}{x(x-y)}\)
= y
\(\frac{x^2 y-x y^2}{x^2-x y}\) = y
3. \(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\)
Solution:
Given
= \(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\)
= \(\frac{(p+q)(p-q)}{(x+y)} \times \frac{(x+y)(x-y)}{(p-q)}\)
= (p+q)(x-y)
\(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\) = (p+q)(x-y)
Solution:
Algebraic Expressions Exercise
Question 1. Let’s see the relations below and find which one is true and which one is false :
1. \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)
Solution:
L.H.S = \(\)
= \(\frac{a+b}{c}\)
= R.H.S
∴ True
2. \(\frac{a}{x+y}=\frac{a}{x}+\frac{a}{y}\)
Solution:
L.H.S = \(\frac{a}{x+y}\)
R.H.S = \(\frac{a}{x}+\frac{a}{y}=\frac{a y+a x}{x y}\)
L.H.S≠R.H.S
∴ False
3. \(\frac{x-y}{a-b}=\frac{y-x}{b-a}\)
Solution:
= \(\frac{x-y}{a-b}\)
= \(\frac{-(x-y)}{-(a-b)}\)
= \(\frac{x-y}{a-b}\)
∵ L.H.S=R.H.S
∴ True
4. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y}\)
Solution:
L.H.S = \(\frac{1}{x}+\frac{1}{y}\)
= \(\frac{y+x}{x y}\)
R.H.S = \(\frac{1}{x+y}\)
∵ L.H.S≠R.H.S
∴ False
Question 2. Let’s express the following algebraic fractions in reduced form:
Solution:
1. \(\frac{63 a^3 b^4}{77 b^5}\)
Solution:
= \(\frac{63 a^3 b^4}{77 b^5}\)
= \(\frac{63 a^3 b^4}{77 b^4}{\prime\cdot b}\)
= \(\frac{9 a^3}{11 b}\)
\(\frac{63 a^3 b^4}{77 b^5}\) = \(\frac{9 a^3}{11 b}\)
2. \(\frac{18 a^4 b^5 c^2}{21 a^7 b^2}\)
Solution:
= \(\frac{18 a^4 b^5 c^2}{77 a^7 b^2}\)
= \(\frac{18 \times a^4 \times b^2 \times b^3 \times c^2}{21 \times a^4 \times a^3 \times b^2}\)
= \(\frac{6 b^3 c^2}{7 a^3}\)
\(\frac{18 a^4 b^5 c^2}{21 a^7 b^2}\) = \(\frac{6 b^3 c^2}{7 a^3}\)
3. \(\frac{x^2-3 x+2}{x^2-1}\)
Solution:
= \(\frac{x^2-3 x+2}{x^2-1}\)
= \(\frac{x^2-2 x-x+2}{(x)^2-(1)^2}\)
= \(\frac{x(x-2)-1(x-2)}{(x+1)(x-1)}\)
= \(\frac{(x-2)(x-1)}{(x+1)(x-1)}\)
= \( \frac{x-2}{x+1}\)
\(\frac{x^2-3 x+2}{x^2-1}\) = \( \frac{x-2}{x+1}\)
4. \(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\)
Solution:
= \(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\)
= \(\frac{a+1}{a-2} \times \frac{a^2-2 a+a-2}{a(a+1)}\)
= \( \frac{(a+1)}{(a-2)} \times \frac{a(a-2)+1(a-2)}{a(a+1)}\)
= \(\frac{(a+1)}{(a-2)} \times \frac{(a-2)(a+1)}{a(a+1)}\)
= \(\frac{a+1}{a}\)
\(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\) = \(\frac{a+1}{a}\)
5. \(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\)
Solution:
= \(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\)
= \(\frac{(p+q)\left(p^2-p q+q^2\right)}{(p+q)(p-q)} \times \frac{(p-q)}{(p+q)}\)
= \(\frac{p^2-p q+q^2}{p+q}\)
\(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\) = \(\frac{p^2-p q+q^2}{p+q}\)
6. \(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\)
Solution:
= \(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\)
= \(\frac{x^2-3 x+2 x-6}{x^2+5 x-x-5} \times \frac{x^2+5 x+x+5}{x^2-3 x-x+3}\)
= \(\frac{x(x-3)+2(x-3)}{x(x+5)-1(x+5)} \times \frac{x(x+5)+1(x+5)}{x(x-3)-1(x-3)}\)
= \(\frac{(x-3)(x+2)}{(x+5)(x-1)} \times \frac{(x+5)(x+1)}{(x-3)(x-1)}\)
= \(\frac{(x+2)(x+1)}{(x-1)(x-1)}\)
\(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\) = \(\frac{(x+2)(x+1)}{(x-1)(x-1)}\)
7. \(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\)
Solution:
= \(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\)
= \(\frac{a^2-a b+b^2}{a(a+b)} \div \frac{(a+b)\left(a-a b+b^2\right)}{(a+b)(a-b)}\)
= \(\frac{\left(a^2-a b+b^2\right)}{a(a+b)} \times \frac{(a+b)(a-b)}{(a+b)\left(a^2-a b+b^2\right)}\)
= \(\frac{a-b}{a}(a+b)
= [latex]\frac{a-b}{a^2+a b}\)
\(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\) = \(\frac{a-b}{a^2+a b}\)
Question 3. Let’s simplify the following algebraic expressions:
Solution:
= \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\)
= \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\)
= \(\frac{c+a+b}{a b c}\)
= \(\frac{a+b+c}{a b c}\)
2. \(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)|
Solution:
= \(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)
= \(\frac{a-\not b-\not c+a+\not b+\not c}{a}\)
= \(\frac{2 a}{a}\)
= 2
\(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)| = 2
3. \(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\)
Solution:
= \(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\)
= \(\frac{x\left(x^2+a^2\right)+b(x-a)-a x^4}{a b x}\)
= \(\frac{x^3+a^2 x+b x-a b-a x^4}{a b x}\)
\(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\) = \(\frac{x^3+a^2 x+b x-a b-a x^4}{a b x}\)
4. \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\)
Solution:
= \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\)
= \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \times \frac{9 a^2}{4 b c^3}\)
= \(\frac{a}{2 b^2}\)
\(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\) = \(\frac{a}{2 b^2}\)
5. \(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\)
Solution:
= \(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\)
= \(\frac{1}{x^2-2 x-x+2}+\frac{1}{x^2-3 x-2 x+6}+\frac{1}{x^2-3 x-x+3}\)
= \(\frac{1}{x(x-2)-1(x-2)}+\frac{1}{x(x-3)-2(x-3)}+\frac{1}{x(x-3)-1(x-3)}\)
= \(\frac{1}{(x-2)(x-1)}+\frac{1}{(x-3)(x-2)}+\frac{1}{(x-3)(x-1)}\)
= \(\frac{x-3+x-1+x-2}{(x-1)(x-2)(x-3)}\)
= \(\frac{3 x-6}{(x-1)(x-2)(x-3)}\)
= \(\frac{3(x-2)}{(x-1)(x-2)(x-3)}\)
= \(\frac{3}{(x-1)(x-3)}\)
=\(\frac{3}{x^2-4 x+3}\)
\(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\) =\(\frac{3}{x^2-4 x+3}\)
6. \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)
Solution:
= \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)
= \(\frac{x+1+x-1}{(x-1)(x+1)}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)
= \(\frac{2 x}{x^2-1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)
= \(\frac{2 x\left(x^2+1\right)+2 x\left(x^2-1\right)}{\left(x^2 1\right)\left(x^2+1\right)}+\frac{4 x^3}{x^4+1}\)
= \(\frac{2 x^3+2 x+2 x^3-2 x}{x^4-1}+\frac{4 x^3}{x^4+1}\)
= \(\frac{4 x^3}{x^4+1}+\frac{4 x^3}{x^4-1}\)
= \(\frac{4 x^3\left(x^4+1\right)+4 x^3\left(x^4-1\right)}{\left(x^4 1\right)\left(x^4+1\right)}\)
= \(\frac{4 x^7+4 x^2+4 x^7-4 x^8}{x^8-1}\)
= \(\frac{8 x^7}{x^8-1}\)
\(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\) = \(\frac{8 x^7}{x^8-1}\)
7. \(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\)
Solution:
= \(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\)
= \(\frac{b(b-5)}{3 b-4 a} \times \frac{(3 b)^2-(4 a)^2}{(b)^2-(5)^2} \div \frac{b(3 b+4 a)}{a(b+5)}\)
= \(\frac{b(b-5)}{(3 b-4 a)} \times \frac{(3 b+4 a)(3 b-4 a)}{(b+5)(b-5)} \times \frac{a(b+5)}{b(3 b+4 a)}\)
= a
\(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\) = a
8. \(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\)
Solution:
= \(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\)
= \(\frac{b+c}{-(a-b)(c-a)}+\frac{c+a}{-(a-b)(b-c)}+\frac{a+b \cdot}{-(c-a)(b-c)}\)
= \( -\left\{\frac{b+c}{(a-b)(c-a)}+\frac{c+a}{(a-b)(b-c)}+\frac{a+b}{(c-a)(b-c)}\right\}\)
= \(-\left\{\frac{(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)}{(a-b)(b-c)(c-a)}\right\}\)
= \(-\left\{\frac{b^2-e^2+e^2-a^2+a^2-b^2}{(a-b)(b-c)(c-a)}\right\}\)
= \(\frac{0}{(a-b)(b-c)(c-a)}=0\)
\(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\) = 0
9. \(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)
Solution:
= \(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)
= \(\frac{b+c-a}{-(a-b)(c-a)}+\frac{c+a-b}{-(b-c)(a-b)}+\frac{a+b-c}{-(c-a)(b-c)}\)
= \(-\left\{\frac{b+c-a}{(a-b)(c-a)}+\frac{c+a-b}{(b-c)(a-b)}+\frac{a+b-c}{(c-a)(b-c)}\right\} \)
= \(-\left\{\frac{(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)}{(a-b)(b-c)(c-a)}\right\}\)
= \(-\left\{\frac{b^2+b c-a b-b c-c^2+a c+c^2+a c-b c-a c-a^2+a b+a^2+a b-a c-a b-b^2+b c}{(a-b)(b-c)(c-a)}\right\}\)
= \(-\frac{0}{(a-b)(b-c)(c-a)}=0\)
\(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)= 0
10. \(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)
Solution:
= \(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)
= \(\frac{\frac{a^2+a x-a^2}{x-a}+\frac{b^2+b x-b^2}{x-b}+\frac{c^2+c x-c^2}{x-c}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)
= \(\frac{\frac{a x}{x-a}+\frac{b x}{x-b}+\frac{c x}{x-c}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)
= \(=\frac{x\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right)}{\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right)}=x\)
\(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)=x
11. \(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)
Solution:
= \(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)
= \(\left\{\frac{\left(a^2+b^2\right)^2-\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)\left(a^2-b^2\right)}\right\} \div\left\{\frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)}\right\} \times\left(\frac{a^2+b^2}{a b}\right)\)
= \(\frac{a^4+2 a^2 b^2+b^4-\left(a^4-2 a^2 b^2+b^4\right)}{\left(a^2+b^2\right)(a+b)(a-b)} \div \frac{a^2+2 a b+b^2-\left(a^2-2 a b+b^2\right)}{(a+b)(a-b)} \times \frac{\left(a^2+b^2\right)}{a b}\)
= \(\frac{a^4+2 a^2 b^2+b^4-a^4+2 a^2 b^2+b^4}{\left(a^2+b^2\right)(a+b)(a-b)} \div \frac{a^2+2 a b+b^2-a^2+2 a b-b^2}{(a+b)(a-b)} \times \frac{\left(a^2+b^2\right)}{a b}\)
= \(=\frac{4 a^2 b^2}{\left(a^2+b^2\right)(a+b)(a-b)} \times \frac{(a+b)(a-b)}{4 a b} \times \frac{\left(a^2+b^2\right)}{a b}\)
= \(\frac{4 a^2 b^2}{4 a^2 b^2}\)
= 1
\(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)=1
12. \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)
Solution:
= \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)
= \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)
= \(-\left\{\frac{y^2+y z+z^2}{(x-y)(z-x)}+\frac{z^2+z x+x^2}{(y-z)(x-y)}+\frac{x^2+x y+y^2}{(z-x)(y-z)}\right\}\)
= \(-\left\{\frac{(y-z)\left(y^2+y z+z^2\right)+(z-x)\left(z^2+z x+x^2\right)+(x-y)\left(x^2+x y+y^2\right)}{(x-y)(y-z)(z-x)}\right\}\)
= \(\frac{y^3-z^3+z^3-x^3+x^3-y^3}{(x-)(y-z)(z-x)}\)
= \(-\frac{0}{(x-y)(y-z)(z-x)}\)
= 0
\(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\) = 0