Đáp án: $C$
Giải thích các bước giải:
Đặt $a^{\frac{1}{2}}=x\rightarrow a=x^2$
$\rightarrow M=\Big(\dfrac{x+2}{x^2+2x+1}-\dfrac{x-2}{x^2-1}\Big):\dfrac{x+1}{x}$
$\rightarrow M=\Big(\dfrac{x+2}{(x+1)^2}-\dfrac{x-2}{(x-1)(x+1)}\Big):\dfrac{x+1}{x}$
$\rightarrow M=\Big(\dfrac{(x+2)(x-1)}{(x+1)^2.(x-1)}-\dfrac{(x-2).(x+1)}{(x-1)(x+1)^2}\Big):\dfrac{x+1}{x}$
$\rightarrow M=\Big(\dfrac{x^2+x-2}{(x+1)^2.(x-1)}-\dfrac{x^2-x-2}{(x-1)(x+1)^2}\Big).\dfrac{x}{x+1}$
$\rightarrow M=\dfrac{x^2+x-2-(x^2-x-2)}{(x+1)^2.(x-1)}.\dfrac{x}{x+1}$
$\rightarrow M=\dfrac{2x}{(x+1)^2.(x-1)}.\dfrac{x}{x+1}$
$\rightarrow M=\dfrac{2}{(x+1).(x-1)}.$
$\rightarrow M=\dfrac{2}{x^2-1}.$
$\rightarrow M=\dfrac{2}{a-1}.$
