$P=\dfrac{x^2+x}{x^2-2x+1}:(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x})\,\,\,\,\,(x\neq1;x\neq0)$
$P=\dfrac{x(x+1)}{(x-1)^2}:(\dfrac{(x+1)(x-1)}{x(x-1)}+\dfrac{x}{x(x-1)}+\dfrac{2-x^2}{x(x-1)})$
$P=\dfrac{x(x+1)}{(x-1)^2}:\dfrac{x^2-1+x+2-x^2}{x(x-1)}$
$P=\dfrac{x(x+1)}{(x-1)^2}\cdot\dfrac{x(x-1)}{x+1}$
$P=\dfrac{x^2}{x-1}$
Để $P=(x-1)^3:x^2$
$⇔\dfrac{x^2}{x-1}=\dfrac{(x-1)^3}{x^2}$
$⇔x^4=(x-1)^4$
$⇔x^4-(x-1)^4=0$
$⇔(x^2-x^2+2x-1)(x^2+x^2-2x+1)=0$
$⇔(2x-1)(2x^2-2x+1)=0$ (1)
Do $2x^2-2x+1=2(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4})+\dfrac{1}{2}=2(x-\dfrac{1}{2})^2+\dfrac{1}{2}≥\dfrac{1}{2}\neq0$
Nên từ (1) $⇒2x-1=0⇔x=\dfrac{1}{2}(tmdk)$
Vậy để $P=(x-1)^3:x^2$ thì $x=\dfrac{1}{2}$
