a) $P = \bigg(\dfrac{1}{x-2}-\dfrac{x^2}{8-x^3} . \dfrac{x^2+2x+4}{x+2}\bigg):\dfrac{1}{x^2-4}$
$ĐKXĐ : x \neq 2, x \neq -2$
Ta có : $P = \bigg(\dfrac{1}{x-2}-\dfrac{x^2}{8-x^3} . \dfrac{x^2+2x+4}{x+2}\bigg):\dfrac{1}{x^2-4}$
$ = \bigg(\dfrac{1}{x-2}+\dfrac{x^2}{x^3-8}.\dfrac{x^2+2x+4}{x+2}\bigg).(x^2-4)$
$ = \bigg(\dfrac{1}{x-2}+\dfrac{x^2}{(x-2).(x^2+2x+4)}.\dfrac{x^2+2x+4}{x+2}\bigg).(x^2-4)$
$ = \bigg(\dfrac{1}{x-2}+\dfrac{x^2}{(x-2).(x+2)}\bigg).(x^2-4)$
$ = \dfrac{x^2+x+2}{x^2-4}.(x^2-4) = x^2+x+2$
Vậy $P=x^2+x+2$ với $x \neq 2, x \neq -2$
b) Ta có $P = x^2+x+2=\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{7}{4} ≥ \dfrac{7}{4}$
Dấu "=" đat được khi $x=-\dfrac{1}{2}$
Vậy Min $P = \dfrac{7}{4}$ khi $x=-\dfrac{1}{2}$
c) Có $x^2+x+2 \vdots x^2+1$
$\to x+1 \vdots x^2+1$
$\to (x-1).(x+1) \vdots x^2+1$
$\to x^2+1-2 \vdots x^2+1$
$\to 2 \vdots x^2+1$
Mà $x^2+1 ≥ 1$ nên $x^2+1 ∈ \bigg\{1,2\bigg\}$
$\to x^2 \in \bigg{0,1\bigg\}$
$\to x \in \{0,-1,1\bigg\}$ ( T/m ĐKXĐ )
