Rút gọn biểu thức C.

Trả lời

Điều kiện xác định: $a>0,a\ne 1$ .

$$\begin{gathered} C=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right) \\ =\left[\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right] \\ =\frac{\sqrt{a}.\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)} \\ =\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)} \\ =\frac{\sqrt{a}+1}{\sqrt{a}}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{\sqrt{a}+1}{\sqrt{a}}.\left(\sqrt{a}-1\right) \\ =\frac{a-1}{\sqrt{a}} \end{gathered}$$

Vậy $C=\frac{a-1}{\sqrt{a}}$  với $a>0,a\ne 1$ .