Đáp án:
\(\begin{array}{l}
a)A = \dfrac{{2\sqrt a + 2a + 2}}{{\sqrt a }}\\
b)\left[ \begin{array}{l}
a = 4\\
a = \dfrac{1}{4}
\end{array} \right.\\
c)a \ne 1;a > 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne 1\\
A = \dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}} + \left( {\dfrac{{a - 1}}{{\sqrt a }}} \right).\dfrac{{a + 2\sqrt a + 1 + a - 2\sqrt a + 1}}{{a - 1}}\\
= \dfrac{{a + \sqrt a + 1 - \left( {a - \sqrt a + 1} \right)}}{{\sqrt a }} + \dfrac{{2a + 2}}{{\sqrt a }}\\
= \dfrac{{2\sqrt a + 2a + 2}}{{\sqrt a }}\\
b)A = 7\\
\to \dfrac{{2\sqrt a + 2a + 2}}{{\sqrt a }} = 7\\
\to 2\sqrt a + 2a + 2 = 7\sqrt a \\
\to 2a - 5\sqrt a + 2 = 0\\
\to \left( {\sqrt a - 2} \right)\left( {2\sqrt a - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt a - 2 = 0\\
2\sqrt a - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
a = 4\\
a = \dfrac{1}{4}
\end{array} \right.\\
c)A > 6\\
\to \dfrac{{2\sqrt a + 2a + 2}}{{\sqrt a }} > 6\\
\to \dfrac{{2\sqrt a + 2a + 2 - 6\sqrt a }}{{\sqrt a }} > 0\\
\to 2a - 4\sqrt a + 2 > 0\left( {do:\sqrt a > 0\forall a > 0} \right)\\
\to a - 2\sqrt a + 1 > 0\\
\to {\left( {\sqrt a - 1} \right)^2} > 0\\
\to a \ne 1;a > 0
\end{array}\)
