X-12x-1-2x-1x-1-52-x - -12-1

Given, x 12x+1+2x+1x 1=52 2(x 1)^2+2(2x+1)^2=5(x 1)(2x+1) 10x^2+4x+4=5(2x^2 x 1) 10x^2+4x+4=10x^2 5x 5 9x= 9 ∴ x= 1 ...

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Byju's Answer

Standard VIII

Mathematics

Solving Equations with Variables on Both Sides

x-12x+1+2x+1x...

Question

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}, x \neq - \frac{1}{2}, 1$

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Solution

Given,

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}$

$2 {(x - 1)}^{2} + 2 {(2 x + 1)}^{2} = 5 (x - 1) (2 x + 1)$

$10 x^{2} + 4 x + 4 = 5 (2 x^{2} - x - 1)$

$10 x^{2} + 4 x + 4 = 10 x^{2} - 5 x - 5$

$9 x = - 9$

$∴ x = - 1$

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Similar questions

Q. Solve :

\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2

, where

x \neq - \frac{1}{2}, 1

Q. Solve for

x \frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2

where

x \neq \frac{- 1}{2}, 1

Q. Solve the following quadratic equation by factorization, the root is :

- 1

and

1

\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}, x \neq - \frac{1}{2}, 1

Solve each of the following quadratic equations:
(i) $\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1$
(ii) $\frac{x - 1}{2 x - 1} + \frac{2 x - 1}{x - 1} = 2, x \neq \frac{1}{2}, 1$

\int \sqrt{(x^{2} + 2 x + 5)} d x

is equal to

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x−12x+1+2x+1x−1=52,x≠−12,1

Given,x−12x+1+2x+1x−1=522(x−1)2+2(2x+1)2=5(x−1)(2x+1)10x2+4x+4=5(2x2−x−1)10x2+4x+4=10x2−5x−59x=−9∴x=−1

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}, x \neq - \frac{1}{2}, 1$

Given,

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}$

$2 {(x - 1)}^{2} + 2 {(2 x + 1)}^{2} = 5 (x - 1) (2 x + 1)$

$10 x^{2} + 4 x + 4 = 5 (2 x^{2} - x - 1)$

$10 x^{2} + 4 x + 4 = 10 x^{2} - 5 x - 5$

$9 x = - 9$

$∴ x = - 1$