Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
\sqrt a - 1 \ne 0\\
\sqrt a \ne 0\\
\sqrt a - 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1\\
a \ne 4
\end{array} \right.\\
b,\\
A = \left( {\dfrac{1}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a - 2}} - \dfrac{{\sqrt a + 2}}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\sqrt a - \left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\sqrt a }}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{1}{{\left( {\sqrt a - 1} \right)\sqrt a }}:\dfrac{{\left( {a - 1} \right) - \left( {a - 4} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{1}{{\left( {\sqrt a - 1} \right)\sqrt a }}:\dfrac{3}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{1}{{\left( {\sqrt a - 1} \right).\sqrt a }}.\dfrac{{\left( {\sqrt a - 2} \right).\left( {\sqrt a - 1} \right)}}{3}\\
= \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
c,\\
A = \dfrac{1}{9} \Leftrightarrow \dfrac{{\sqrt a - 2}}{{3\sqrt a }} = \dfrac{1}{9}\\
\Leftrightarrow 9.\left( {\sqrt a - 2} \right) = 3\sqrt a \\
\Leftrightarrow 9\sqrt a - 18 = 3\sqrt a \\
\Leftrightarrow 6\sqrt a = 18\\
\Leftrightarrow \sqrt a = 3\\
\Leftrightarrow a = 9\\
d,\\
a = 9 + 4\sqrt 5 = 5 + 2.\sqrt 5 .2 + 4 = {\left( {\sqrt 5 + 2} \right)^2}\\
\Rightarrow \sqrt a = \sqrt 5 + 2\\
A = \dfrac{{\sqrt a - 2}}{{3\sqrt a }} = \dfrac{{\left( {\sqrt 5 + 2} \right) - 2}}{{3.\left( {\sqrt 5 + 2} \right)}} = \dfrac{{\sqrt 5 }}{{3.\left( {\sqrt 5 + 2} \right)}} = \dfrac{{\sqrt 5 .\left( {\sqrt 5 - 2} \right)}}{{3.\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)}} = \dfrac{{5 - 2\sqrt 5 }}{{3.\left( {5 - 4} \right)}} = \dfrac{{5 - 2\sqrt 5 }}{3}
\end{array}\)
