Đáp án:
$A = \sqrt a - a$ với ${a \ge 0;a \ne 1}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = \left( {\dfrac{{\sqrt a - 2}}{{a - 1}} - \dfrac{{\sqrt a + 2}}{{a + 2\sqrt a + 1}}} \right).\dfrac{{{a^2} - 2a + 1}}{2}\left( {DK:a \ge 0;a \ne 1} \right)\\
= \left( {\dfrac{{\sqrt a - 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} - \dfrac{{\sqrt a + 2}}{{{{\left( {\sqrt a + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {a - 1} \right)}^2}}}{2}\\
= \left( {\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right){{\left( {\sqrt a + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {a - 1} \right)}^2}}}{2}\\
= \dfrac{{ - 2\sqrt a }}{{\left( {\sqrt a - 1} \right){{\left( {\sqrt a + 1} \right)}^2}}}.\dfrac{{{{\left( {a - 1} \right)}^2}}}{2}\\
= \dfrac{{ - 2\sqrt a }}{{\left( {\sqrt a - 1} \right){{\left( {\sqrt a + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}{{\left( {\sqrt a + 1} \right)}^2}}}{2}\\
= - \sqrt a \left( {\sqrt a - 1} \right)\\
= \sqrt a - a
\end{array}$
Vậy $A = \sqrt a - a$ với ${a \ge 0;a \ne 1}$
