Rút gọn B

Trả lời

Điều kiện: $x\ge 0,x\ne 1$ .

$$\begin{gathered} B=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}+\frac{6\sqrt{x}-4}{1-x} \\ =\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ ==\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{{\left(\sqrt{x}-1\right)}^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)} \\ =\frac{\sqrt{x}-1}{\sqrt{x}+1} \end{gathered}$$

Vậy $B=\frac{\sqrt{x}-1}{\sqrt{x}+1}$  với $x\ge 0,x\ne 1$ .