a) $P=\bigg(\dfrac{x^3+1}{x+1}-x\bigg):(x-1)$ ĐK: $x\neq\pm1$
$P=\bigg(\dfrac{(x+1)(x^2-x+1)}{x+1}-\dfrac{x(x+1)}{x+1}\bigg):(x-1)$
$P=\dfrac{(x+1)(x^2-x+1)-x(x+1)}{x+1}.\dfrac{1}{x-1}$
$P=\dfrac{(x+1)(x^2-x+1-x)}{x+1}.\dfrac{1}{x-1}$
$P=(x^2-2x+1).\dfrac{1}{x-1}$
$P=\dfrac{(x-1)^2}{x-1}$
$P=x-1$
Vậy $P=x-1$ với $x\neq\pm1$
b) Để $P=x^2-7$ ĐK: $x\neq\pm1$
$⇔x-1=x^2-7$
$⇔x^2-x-6=0$
$⇔(x-3)(x+2)=0$
\(⇔\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=3(tm)\\x=-2(tm)\end{array} \right.\)
Vậy với $x=3$ hoặc $x=-2$ thì $P=x^2-7$