Đáp án:
a) \(\dfrac{{2a + 2\sqrt a + 2}}{{\sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne 1\\
A = \dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}} + \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{{a + 2\sqrt a + 1 + a - 2\sqrt a + 1}}{{a - 1}}\\
= \dfrac{{a + \sqrt a + 1 - a + \sqrt a - 1}}{{\sqrt a }} + \dfrac{{2a + 2}}{{\sqrt a }}\\
= \dfrac{{2a + 2\sqrt a + 2}}{{\sqrt a }}\\
b)A > 6\\
\to \dfrac{{2a + 2\sqrt a + 2}}{{\sqrt a }} > 6\\
\to \dfrac{{2a + 2\sqrt a + 2 - 6\sqrt a }}{{\sqrt a }} > 0\\
\to 2a - 4\sqrt a + 2 > 0\left( {do:\sqrt a > 0\forall a > 0} \right)\\
\to a - 2\sqrt a + 1 > 0\\
\to {\left( {\sqrt a - 1} \right)^2} > 0\\
\to a \ne 1
\end{array}\)
