Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0\\
A = \dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} - \dfrac{{2a + \sqrt a }}{{\sqrt a }} + 1\\
= \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - \dfrac{{\sqrt a \left( {2\sqrt a + 1} \right)}}{{\sqrt a }} + 1\\
= \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - 2\sqrt a - 1 + 1\\
= \sqrt a \left( {\sqrt a + 1} \right) - 2\sqrt a \\
= a + \sqrt a - 2\sqrt a \\
= a - \sqrt a \\
b)a > 1\\
\Rightarrow A = \sqrt a \left( {\sqrt a - 1} \right) > 0\\
\Rightarrow A = \left| A \right|\\
c)A = 2\\
\Rightarrow a - \sqrt a = 2\\
\Rightarrow a - \sqrt a - 2 = 0\\
\Rightarrow \left( {\sqrt a - 2} \right)\left( {\sqrt a + 1} \right) = 0\\
\Rightarrow \sqrt a = 2\\
\Rightarrow a = 4\left( {tmdk} \right)\\
Vay\,a = 4\\
d)A = a - \sqrt a \\
= a - 2.\sqrt a .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {\sqrt a - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\Rightarrow A \ge - \dfrac{1}{4}\\
\Rightarrow GTNN:A = - \dfrac{1}{4}\\
Khi:\sqrt a = \dfrac{1}{2}\\
\Rightarrow a = \dfrac{1}{4}\left( {tmdk} \right)
\end{array}$
