Solve this

Given :

$\left(\frac{x}{x+1}\right)^{2}-5\left(\frac{x}{x+1}\right)+6=0$

Putting $\frac{x}{x+1}=y$, we get:

$y^{2}-5 y+6=0$

$\Rightarrow y^{2}-5 y+6=0$

$\Rightarrow y^{2}-(3+2) y+6=0$

$\Rightarrow y^{2}-3 y-2 y+6=0$

$\Rightarrow y(y-3)-2(y-3)=0$

$\Rightarrow(y-3)(y-2)=0$

$\Rightarrow y-3=0$ or $y-2=0$

$\Rightarrow y=3$ or $y=2$

Case I

If $y=3$, we get:

$\frac{x}{x+1}=3$

$\Rightarrow x=3(x+1)[$ On cross multiplying $]$

$\Rightarrow x=3 x+3$

$\Rightarrow x=\frac{-3}{2}$

Case II

If $y=2$, we get:

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

$\Rightarrow x=2(x+1)$

$\Rightarrow x=2 x+2$

$\Rightarrow-x=2$

$\Rightarrow x=-2$

Hence, the roots of the equation are $\frac{-3}{2}$ and $-2$.