Đáp án:
Giải thích các bước giải:
$M=\left[\dfrac{1}{1+\dfrac{1}{3}\left(2\sqrt{x}+1\right)^2}+\dfrac{1}{1+\dfrac{1}{3}\left(2\sqrt{x}-1\right)^2}\right]\cdot \dfrac{1}{x+1}$
$=\dfrac{1}{x+1}\left(\dfrac{3}{4x+4\sqrt{x}+4}+\dfrac{3}{4x-4\sqrt{x}+4}\right)$
$=\dfrac{3\left(x+1\right)}{2\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot \dfrac{1}{x+1}$
$=\dfrac{3\left(x+1\right)}{2\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\left(x+1\right)}$
$=\dfrac{3}{2\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}$
`=(1)/(2x^2 + 2x+2)`
