Cho biểu thức ​\(B=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2 \sqrt{x}+1}\right) \cdot \frac{(1-x)^{2}}{2} \text { với } x \geq 0 ; x \neq 1\).Rút gọn biểu thức B ta được:

* Hướng dẫn giải

 \(\begin{array}{l} B = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right) \cdot \frac{{{{(1 - x)}^2}}}{2} = \left( {\frac{{\sqrt x - 2}}{{(\sqrt x - 1)(\sqrt x + 1)}} - \frac{{\sqrt x + 2}}{{{{(\sqrt x + 1)}^2}}}} \right) \cdot \frac{{{{(x - 1)}^2}}}{2}\\ = \left( {\frac{{(\sqrt x - 2)(\sqrt x + 1)}}{{(\sqrt x - 1){{(\sqrt x + 1)}^2}}} - \frac{{(\sqrt x + 2)(\sqrt x - 1)}}{{(\sqrt x - 1){{(\sqrt x + 1)}^2}}}} \right) \cdot \frac{{{{(\sqrt x - 1)}^2}{{(\sqrt x + 1)}^2}}}{2}\\ = \frac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{(\sqrt x - 1){{(\sqrt x + 1)}^2}}} \cdot \frac{{{{(\sqrt x - 1)}^2} \cdot {{(\sqrt x + 1)}^2}}}{2} = \frac{{ - 2\sqrt x (\sqrt x - 1)}}{2} = \sqrt x - x \end{array}\)