Question

answer this question​

Answer 1

Explanation:

$\mathsf{Let\: \frac{x}{x+1}\:be\:y.}$

$\mathsf{5(\frac{x}{x+1})^2-4(\frac{x}{x+1})-1=0}$

$\mathsf{==>5y^2-4y-1=0}$

$\mathsf{==>5y^2-5y+y-1=0}$

$\mathsf{==>5y(y-1)+1(y-1)=0}$

$\mathsf{==>(y-1)(5y+1)=0}$

∴ $\mathsf{y=1\:OR\:y=\frac{-1}{5}}$

But,

$\mathsf{y=\frac{x}{x+1}}$, meaning

$\mathsf{\frac{x}{x+1}=1 ==>x=x+1}$ which does not give us a clear solution.

So we will check the other solution.

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

$\mathsf{==>5x=-(x+1)}$

$\mathsf{==>5x=-x-1}$

$\mathsf{==>6x=-1}$

∴ $\mathsf{x=\frac{-1}{6}}$

Hence, since this gives us a real solution, x = -1/6,

Answer: x = -1/6

I hope this helps. :D

Answer 2

see the attachment

answer is minus one upon six