a) Ta có:
\(Q = \left( {\dfrac{1}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a - 2}} - \dfrac{{\sqrt a + 2}}{{\sqrt a - 1}}} \right)\)
\( = \dfrac{{\sqrt a - \left( {\sqrt a - 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}\)
\( = \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{3}\)
\( = \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\)
b) Ta có: \(a > 0\) nên \(\sqrt a > 0\)
Khi đó: \(Q = \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\) dương khi \(\sqrt a - 2 > 0\)
Ta có: \(\sqrt a - 2 > 0 \Leftrightarrow \sqrt a > 2 \Leftrightarrow a > 4\)
Vậy khi \(a>4\) thì \(Q>0.\)
