Đáp án:
a, $P=\dfrac{2x+1}{x-2}$
b, $P=-3$
c, $x \in \{-3;1;3;7\}$
Giải thích các bước giải:
a, ĐKXĐ: $x \ne \pm 2$
$P=\dfrac1{x-2}+\dfrac{2x^2+4x}{x^2-4}$
$=\dfrac1{x-2}+\dfrac{2x(x+2)}{(x-2)(x+2)}$
$=\dfrac1{x-2}+\dfrac{2x}{x-2}$
$=\dfrac{2x+1}{x-2}$
b, Thay $x=1$ (TMĐKXĐ) vào $P$, ta có:
$P=\dfrac{2x+1}{x-2}=\dfrac{2.1+1}{1-2}=-3$
c, Để $P \in \Bbb Z$ thì $\dfrac{2x+1}{x-2} \in \Bbb Z$
$\to \dfrac{2x-4+5}{x-2} \in \Bbb Z$
$\to \dfrac{2(x-2)+5}{x-2} \in \Bbb Z$
$\to 2+\dfrac5{x-2} \in \Bbb Z$
$\to \dfrac5{x-2} \in \Bbb Z$
$\to 5\ \vdots\ x-2$
$\to x-2 \in Ư(5)$
$\to x-2 \in \{-5;-1;1;5\}$
$\to x \in \{-3;1;3;7\}$
Vậy $x \in \{-3;1;3;7\}$ thì $P$ có giá trị nguyên.
