Đáp án:
$A=\dfrac{x+x\sqrt{x}-3}{(x\sqrt{x}-1)(x+1)}$
Giải thích các bước giải:
$A=\dfrac{x-2\sqrt{x}}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}+1}+\dfrac{2x-2\sqrt{x}}{x^2-\sqrt{x}}$ ĐK: $x\neq\pm1;x>0$
$A=\dfrac{x-2\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{\sqrt{x}+1}{(x+1)\sqrt{x}+(x+1)}+\dfrac{2x-2\sqrt{x}}{\sqrt{x}(x\sqrt{x}-1)}$
$A=\dfrac{x-2\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{\sqrt{x}+1}{(x+1)(\sqrt{x}+1)}+\dfrac{2\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}$
$A=\dfrac{x-2\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{1}{x+1}+\dfrac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$
$A=\dfrac{x-2\sqrt{x}+2\sqrt{x}-2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{1}{x+1}$
$A=\dfrac{x-2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{1}{x+1}$
$A=\dfrac{x-2+x\sqrt{x}-1}{(x\sqrt{x}-1)(x+1)}$
$A=\dfrac{x+x\sqrt{x}-3}{(x\sqrt{x}-1)(x+1)}$
Vậy $A=\dfrac{x+x\sqrt{x}-3}{(x\sqrt{x}-1)(x+1)}$ với $x\neq\pm1;x>0$
