Question

plz solve it fast....with quadritic method... plz....faltu ke ans die na toh dekh lena​

Answer 1

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

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

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

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

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

$\frac{-3}{x^{2} -1}$=$\frac{-1}{8}$

-24=-$x^{2}$+1

$x^{2}$-24-1=0

$x^{2}$-25=0

Now,using Quadratic formula

D=b^2 -4ac=(0)^2 -4(1)(-25)

=100

x= -b ±√D/2a

= -(0)±√100/2(1)

= ±10/2

=±5

Answer 2

Answer:

answer in the above attachment

hope it helps