Đáp án:
$F=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}$
$G=\dfrac{x+4}{\sqrt{x}}$
Giải thích các bước giải:
$F=\left( \dfrac{3\sqrt{x}+5}{x-\sqrt{x}-2}+\dfrac{2\sqrt{x}+4}{4-x} \right):\dfrac{\sqrt{x}}{x+\sqrt{x}}$
$F=\left[ \dfrac{3\sqrt{x}+5}{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-2 \right)}-\dfrac{2\left( \sqrt{x}+2 \right)}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)} \right]:\dfrac{\sqrt{x}}{\sqrt{x}\left( \sqrt{x}+1 \right)}$
$F=\left[ \dfrac{3\sqrt{x}+5}{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-2 \right)}-\dfrac{2}{\sqrt{x}-2} \right]:\dfrac{1}{\sqrt{x}+1}$
$F=\dfrac{3\sqrt{x}+5-2\left( \sqrt{x}+1 \right)}{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-2 \right)}:\dfrac{1}{\sqrt{x}+1}$
$F=\dfrac{\sqrt{x}+3}{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-2 \right)}\cdot\left( \sqrt{x}+1 \right)$
$F=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}$
$G=\left( \dfrac{x\sqrt{x}-x}{x-1}+\dfrac{4\sqrt{x}}{x+\sqrt{x}} \right):\dfrac{\sqrt{x}}{\sqrt{x}+1}$
$G=\left[ \dfrac{x\left( \sqrt{x}-1 \right)}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+1 \right)}+\dfrac{4\sqrt{x}}{\sqrt{x}\left( \sqrt{x}+1 \right)} \right]:\dfrac{\sqrt{x}}{\sqrt{x}+1}$
$G=\left( \dfrac{x}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1} \right):\dfrac{\sqrt{x}}{\sqrt{x}+1}$
$G=\dfrac{x+4}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}$
$G=\dfrac{x+4}{\sqrt{x}}$
