Đáp án: $Q=1$
Giải thích các bước giải:
Ta có:
$Q=\dfrac{x}{xy+x+2}+\dfrac{y}{yz+y+1}+\dfrac{2z}{xz+2z+2}$
$\to Q=\dfrac{x}{xy+x+2}+\dfrac{xy}{xyz+xy+x}+\dfrac{2z}{xz+2z+xyz}$ vì $xyz=2$
$\to Q=\dfrac{x}{xy+x+2}+\dfrac{xy}{2+xy+x}+\dfrac{2}{x+2+xy}$ vì $xyz=2$
$\to Q=\dfrac{x}{xy+x+2}+\dfrac{xy}{xy+x+2}+\dfrac{2}{xy+x+2}$
$\to Q=\dfrac{x+xy+2}{xy+x+2}$
$\to Q=1$