Đáp án:$A=\sqrt{x}-1$
Giải thích các bước giải:
$A=\left ( \frac{2x+1}{x\sqrt{x}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1} \right ).\left ( \frac{x\sqrt{x}+1}{1+\sqrt{x}}-\sqrt{x} \right )\\
=\left ( \frac{2x+1}{(\sqrt{x}-1).(x+\sqrt{x}+1)}-\frac{\sqrt{x}.(\sqrt{x}-1)}{(\sqrt{x}+1).(x+\sqrt{x}+1)} \right ).\left ( \frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}+1}-\sqrt{x} \right )\\
=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.(x-2\sqrt{x}+1)\\
=\frac{1}{\sqrt{x}-1}.(\sqrt{x}-1)^2=\sqrt{x}-1$
