Question

Solve:

$\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x-1}\right)=\frac{7}{8}$

Answer 1

$\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x-1}\right)=\frac{7}{8}$

$\Rightarrow \left(\frac{x+2}{x+1}\right)\left(\frac{x}{x-1}\right)=\frac{7}{8}$

$\Rightarrow \frac{x\left(x+2\right)}{x^2-1}=\frac{7}{8}$

$\Rightarrow \frac{x^2+2x}{x^2-1}=\frac{7}{8}$

$\Rightarrow 8x^2+16x=7x^2-7$

$\Rightarrow x^2+16x+7=0$

$x=\frac{-16\pm \sqrt{256-28}}{2}$

$x=\frac{-16\pm \sqrt{228}}{2}$

$=\frac{-16\pm \sqrt{2^2\times 57}}{2}$

$=-8\pm \sqrt{57}$