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