Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 4;x\# 9\\
A = \left( {\dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} + \dfrac{{\sqrt x + 2}}{{3 - \sqrt x }} - \dfrac{{10 - 5\sqrt x }}{{x - 5\sqrt x + 6}}} \right)\\
:\dfrac{{\sqrt x + 1}}{{\sqrt x \left( {x - 4\sqrt x + 4} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - 10 + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
.\dfrac{{\sqrt x {{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt {x + 1} }}\\
= \dfrac{{x - 4\sqrt x + 3 - x + 4 - 10 + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x {{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 3}}.\dfrac{{\sqrt x }}{{\sqrt {x + 1} }}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)A = 2\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x + 1}} = 2\\
\Leftrightarrow \sqrt x = 2\sqrt x + 2\\
\Leftrightarrow \sqrt x + 2 = 0\left( {ktm} \right)
\end{array}$
Vậy ko có x để A=2
$\begin{array}{l}
A = \dfrac{{\sqrt x }}{{\sqrt x + 1}} \ge 0\left( {do:x \ge 0} \right)\\
\Leftrightarrow GTNN:A = 0\\
Khi:x = 0
\end{array}$
