delta23 wrote:

If x is an integer and \(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\), what is the value of x?

A) -3

B) -1

C) 0

D) 2

E) 3

\(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\)

Or, \(\frac{x(x+1)+4(x-1)}{2(x+1)(x-1)}=\frac{4x(1-x)+(1+x)}{2(1+x)(1-x)}\)

Or, \(\frac{x^2+5x-4}{2(x+1)(x-1)}= \frac{-4x^2+5x+1}{-2(x+1)(x-1)}\)

Or,\(x^2+5x-4=-(-4x^2+5x+1)\)

Or, \(x^2+5x-4=4x^2-5x-1\)

Or, \(3x^2-9x-x+3=0\)

Or, 3x(x-3)-1(x-3)=0

Or, (x-3)(3x-1)=0

Or,

x=3 or, x=\(\frac{1}{3}\)

Given,x is an integer. Hence, x=3.

Ans. (E)

_________________

Regards,

PKN

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