Cho biểu thức $A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right)$.

a) Rút gọn A.

a) Điều kiện xác định: x > 0, x ≠ 1.

Ta có:

$$\begin{gathered} A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right) \\ =\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right] \\ =\left[\frac{\left(x+\sqrt{x}+1\right)}{\sqrt{x}}-\frac{\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right] \\ =\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\frac{x+\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =\frac{2\sqrt{x}}{\sqrt{x}}:\frac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =2.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+1} \\ =\frac{{\displaystyle 2\text{x}-2}}{{\displaystyle x+1}} \end{gathered}$$

Vậy với x > 0, x ≠ 1 thì $A=\frac{2\text{x}-2}{x+1}$.