C= {[(căn x -2) phần (x-1)] - [(căn x +2) phần (x - 2 căn x +1)]}* (1-x) ^2 phần 2

\[\begin{array}{l} C = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x - 2\sqrt x + 1}}} \right).\frac{{{{\left( {1 - x} \right)}^2}}}{2}\\ DK:\,\,\,x \ge 0,\,\,\,x \ne 1\\ \Rightarrow C = \left[ {\frac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \frac{{\sqrt x + 2}}{{{{\left( {\sqrt x - 1} \right)}^2}}}} \right].\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x - 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{x - 3\sqrt x + 2 - x - 3\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {x - 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{ - 6\sqrt x }}{{\sqrt x - 1}}.\frac{{x - 1}}{2} = \frac{{ - 3\sqrt x \left( {x - 1} \right)}}{{\sqrt x - 1}}\\ = \frac{{ - 3\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}} = - 3\sqrt x \left( {\sqrt x + 1} \right). \end{array}\]