Đáp án:
$\begin{array}{l}
a)Đkxđ:x > 0\\
Y = \frac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} - 1 - \frac{{2x + \sqrt x }}{{\sqrt x }}\\
= \frac{{\sqrt x \left( {x\sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - 1 - \frac{{\sqrt x \left( {2\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \frac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - 1 - \left( {2\sqrt x + 1} \right)\\
= \sqrt x \left( {\sqrt x + 1} \right) - 1 - 2\sqrt x - 1\\
= x + \sqrt x - 2\sqrt x - 2\\
= x - \sqrt x - 2\\
b)Y = x - \sqrt x - 2\\
= x - 2.\frac{1}{2}.\sqrt x + \frac{1}{4} - \frac{9}{4}\\
= {\left( {\sqrt x - \frac{1}{2}} \right)^2} - \frac{9}{4} \ge - \frac{9}{4}\forall x > 0\\
\Rightarrow GTNN:Y = - \frac{9}{4} \Leftrightarrow \sqrt x = \frac{1}{2} \Rightarrow x = \frac{1}{4}\\
c)x \ge 4\\
\Rightarrow \sqrt x \ge 2\\
\Rightarrow {\left( {\sqrt x - \frac{1}{2}} \right)^2} = \frac{9}{4}\\
\Rightarrow Y = {\left( {\sqrt x - \frac{1}{2}} \right)^2} - \frac{9}{4} = 0\\
\Rightarrow \left| Y \right| = 0\\
\Rightarrow Y - \left| Y \right| = 0
\end{array}$
