Đáp án:
\[C = - \sqrt x .\left( {\sqrt x - 1} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
C = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\\
= \left( {\dfrac{{\sqrt x - 2}}{{{{\sqrt x }^2} - {1^2}}} - \dfrac{{\sqrt x + 2}}{{{{\sqrt x }^2} + 2.\sqrt x .1 + {1^2}}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \left( {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {{{\sqrt x }^2} - {1^2}} \right)}^2}}}{2}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right).{{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left[ {\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)} \right]}^2}}}{2}\\
= \dfrac{{\left( {x + \sqrt x - 2\sqrt x - 2} \right) - \left( {x - \sqrt x + 2\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= \dfrac{{\left( {x - \sqrt x - 2} \right) - \left( {x + \sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= \dfrac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= \dfrac{{ - 2\sqrt x }}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\
= - \sqrt x .\left( {\sqrt x - 1} \right)
\end{array}\)
