Question

Solve the following inequalities.
$\frac{(x-1{)}^{2}x}{x+1}⩽0$

Answer 1

$\frac{(x-1{)}^{2}x}{x+1}\le 0$

Here, $x\ne -1$

Since, $(x-1{)}^2\ge 0$

$⟹\frac{x}{x+1}\le 0$ so that overall fraction is non-positive.

$⟹-1\le x\le 0$

But, $x\ne -1$

Hence, solution set is-

$x\in (-1,0]$