Đáp án+giải thích các bước giải:
$\text{A=}$ $(\dfrac{x}{x^2-1}+\dfrac{1}{2x-2}+\dfrac{1}{2x+2}).(1+\dfrac{1}{x})$
$\text{ĐKXĐ: x $\neq$ 0; x $\neq$ ± 1}$
a)
$\text{A=}$ $(\dfrac{x}{x^2-1}+\dfrac{1}{2x-2}+\dfrac{1}{2x+2}).(1+\dfrac{1}{x})$
$=(\dfrac{2x}{2(x-1)(x+1)}+\dfrac{x+1}{2(x-1)(x+1)}+\dfrac{x-1}{2(x-1)(x+1)}).\dfrac{x+1}{x}$
$=(\dfrac{2x+x+1+x-1}{2(x-1)(x+1)}).\dfrac{x+1}{x}$
$=\dfrac{3x}{2(x-1)(x+1)}.\dfrac{x+1}{x}$
$=\dfrac{4x.(x+1)}{2x(x-1)(x+1)}$
$=\dfrac{2}{x-1}$
b)
c)
$\text{Để A nguyên (x}$ $\neq 1)$
$\text{⇔2 phải chia hết cho (x-1)}$
$⇔ (x-1) ∈ Ư(2)=${$±1; ±2$}
$⇒$\(\left[ \begin{array}{l}x-1=1\\x-1=-1\\x-1=2\\x-1=-2\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2\\x=0\\x=3\\x=-1\end{array} \right.\)
$\text{Vậy x∈}$ {$-1; 0; 2; 3$} $\text{thì GTBT A nguyên}$
