Đáp án: $Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$
Giải thích các bước giải:
Với $x\ge 0\,,\,x\ne 1$:
$Q=\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}$
$Q=\dfrac{3x-3\sqrt{x}-3}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}-\dfrac{\sqrt{x}+1}{\left( \sqrt{x}+2 \right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}$
$Q=\dfrac{3x-3\sqrt{x}-3-\left( \sqrt{x}+1 \right)\left( \sqrt{x}-1 \right)-\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}$
$Q=\dfrac{3x-3\sqrt{x}-3-\left( x-1 \right)-\left( x-4 \right)}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}$
$Q=\dfrac{3x-3\sqrt{x}-3-x+1-x+4}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}$
$Q=\dfrac{x-3\sqrt{x}+2}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}$
$Q=\dfrac{\left( \sqrt{x}-1 \right)\left( \sqrt{x}-2 \right)}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+2 \right)}$
$Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$
