Tìm đkxđ và rút gọn biểu thức: C= $(\frac{\sqrt[]{x}-2}{x-1}-$ $\frac{\sqrt[]{x}+2}{x+2\sqrt[]{x}+1})$$.\frac{(1-x)^2}{2}$

Điều kiện xác định:

$\begin{array}{l} \left\{ \begin{array}{l} x – 1 \ne 0\\ x \ge 0\\ x + 2\sqrt x  + 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne 1\\ x \ge 0\\ {\left( {\sqrt x  + 1} \right)^2} \ne 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} x \ne 1\\ x \ge 0 \end{array} \right. \end{array}$

$\begin{array}{l} C\left( {\dfrac{{\sqrt x  – 2}}{{x – 1}} – \dfrac{{\sqrt x  + 2}}{{x + 2\sqrt x  + 1}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\ C = \left( {\dfrac{{\sqrt x  – 2}}{{\left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}} – \dfrac{{\sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\ C = \dfrac{{\left( {\sqrt x  – 2} \right)\left( {\sqrt x  + 1} \right) – \left( {\sqrt x  + 2} \right)\left( {\sqrt x  – 1} \right)}}{{\left( {x – 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\ C = \dfrac{{x – \sqrt x  – 2 – x – \sqrt x  + 2}}{{\left( {x – 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\ C = \dfrac{{ – \sqrt x \left( {x – 1} \right)}}{{\sqrt x  + 1}}\\C=\dfrac{-\sqrt x(\sqrt x-1)(\sqrt x+1)}{\sqrt x +1}\\C=-\sqrt{x}(\sqrt x-1) \end{array}$