Giải thích các bước giải:
a) ĐK: $a>0,a\ne 1$
$\begin{array}{l}
\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\
= \left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right)^2}\\
= \left( {a + 2\sqrt a + 1} \right){\left( {\dfrac{1}{{1 + \sqrt a }}} \right)^2}\\
= {\left( {1 + \sqrt a } \right)^2}{\left( {\dfrac{1}{{1 + \sqrt a }}} \right)^2}\\
= 1
\end{array}$
Ta có đpcm
b) ĐKXĐ: $x>0, x\ne 1$
Ta có:
$\begin{array}{l}
\left( {\dfrac{{2 + \sqrt x }}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{x - 1}}} \right).\dfrac{{x\sqrt x + x - \sqrt x - 1}}{{\sqrt x }}\\
= \left( {\dfrac{{2 + \sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\sqrt x - 2}}{{x - 1}}} \right).\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{\left( {2 + \sqrt x } \right)\left( {x - 1} \right) - \left( {\sqrt x - 2} \right){{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x + 2x}}{{\left( {x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}{{\sqrt x }}\\
= 2
\end{array}$
Vậy ta có đpcm
