Đáp án:
$P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}$
$Q=x-\sqrt{x}+1$
Giải thích các bước giải:
$P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\,\,\,\,\,\,\text{ĐK: $x\ge0,x\neq1$}\\P=\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\P=\dfrac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\P=\dfrac{x+\sqrt{x}+1-x-2+x-1}{(\sqrt{x}+1)(x+\sqrt{x}+1)}\\P=\dfrac{x+\sqrt{x}}{(\sqrt{x}+1)(x+\sqrt{x}+1)}\\P=\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(x+\sqrt{x}+1)}\\P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}$
Vậy $P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}$ với $x\ge0,x\neq1$
$Q=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2(x-1)}{\sqrt{x}-1}\,\,\,\,\,\text{ĐK: $x>0,x\neq1$}\\Q=\dfrac{\sqrt{x}(x\sqrt{x}-1)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\dfrac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\\Q=\sqrt{x}(\sqrt{x}-1)-2\sqrt{x}-1+2(\sqrt{x}+1)\\Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\Q=x-\sqrt{x}+1$
Vậy $Q=x-\sqrt{x}+1$ với $x>0,x\neq1$
