Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
a > 0\\
a \ne 1\\
a \ne 4
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
Q = \left( {\frac{1}{{\sqrt a - 1}} - \frac{1}{{\sqrt a }}} \right):\left( {\frac{{\sqrt a + 1}}{{\sqrt a - 2}} - \frac{{\sqrt a + 2}}{{\sqrt a - 1}}} \right)\\
= \left( {\frac{{\sqrt a - \left( {\sqrt a - 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right):\left( {\frac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}} \right)\\
= \frac{1}{{\sqrt a .\left( {\sqrt a - 1} \right)}}:\frac{{\left( {a - 1} \right) - \left( {a - 4} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \frac{1}{{\sqrt a .\left( {\sqrt a - 1} \right)}}.\frac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{3}\\
= \frac{{\sqrt a - 2}}{{3\sqrt a }}
\end{array}\)
b,
\(Q > 0 \Leftrightarrow \frac{{\sqrt a - 2}}{{3\sqrt a }} > 0 \Leftrightarrow \sqrt a - 2 > 0 \Leftrightarrow a > 4\)
c,
\(\begin{array}{l}
a = 9 - 4\sqrt 5 = {2^2} - 2.2.\sqrt 5 + 5 = {\left( {\sqrt 5 - 2} \right)^2}\\
\Rightarrow \sqrt a = \sqrt 5 - 2\\
\Rightarrow Q = \frac{{\sqrt a - 2}}{{3\sqrt a }} = \frac{{\sqrt 5 - 4}}{{3\sqrt 5 - 6}}
\end{array}\)
