Đáp án:
$\begin{array}{l}
a)Dkxd:a \ge 0\\
A = \left( {\dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}}} \right) - \left( {\dfrac{{2a + \sqrt a }}{{\sqrt a + 1}}} \right)\\
= \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - \dfrac{{2a + \sqrt a }}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - \dfrac{{2a + \sqrt a }}{{\sqrt a + 1}}\\
= \sqrt a \left( {\sqrt a + 1} \right) - \dfrac{{2a + \sqrt a }}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a {{\left( {\sqrt a + 1} \right)}^2} - 2a - \sqrt a }}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a \left( {a + 2\sqrt a + 1} \right) - 2a - \sqrt a }}{{\sqrt a + 1}}\\
= \dfrac{{a\sqrt a }}{{\sqrt a + 1}}\\
b)a \ge 1 \Rightarrow \sqrt a \ge 1\\
A = \dfrac{{a\sqrt a }}{{\sqrt a + 1}}\\
\Rightarrow A = \left| A \right|\\
c)Dkxd:a \ge 0\\
A = 2\\
\Rightarrow \dfrac{{a\sqrt a }}{{\sqrt a + 1}} = 2\\
\Rightarrow a\sqrt a = 2\sqrt a + 2\\
\Rightarrow a\sqrt a - 2\sqrt a - 2 = 0\\
\Rightarrow \sqrt a = 1,769\\
\Rightarrow a = 3,13\left( {tmdk} \right)\\
d)A = \dfrac{{a\sqrt a }}{{\sqrt a + 1}} \ge 0\\
\Rightarrow GTNN:A = 0\,khi:a = 0
\end{array}$
