Đáp án:
$\begin{array}{l}
1)Dkxd:\dfrac{{{x^2} + 1}}{{5x - 1}} \ge 0\\
\Leftrightarrow 5x - 1 > 0\left( {do:{x^2} + 1 > 0} \right)\\
\Leftrightarrow 5x > 1\\
\Leftrightarrow x > \dfrac{1}{5}\\
Vậy\,x > \dfrac{1}{5}\\
2)a)Dkxd:\left\{ \begin{array}{l}
x > 0\\
x\# 1
\end{array} \right.\\
Q = \left( {\dfrac{{{x^2}}}{{{x^2} - \sqrt {{x^3}} }} + \dfrac{{x + 1}}{{\sqrt x }} - \dfrac{{\sqrt x + 1}}{{x - 1}}} \right).\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \left( {\dfrac{{{x^2}}}{{x\left( {x - \sqrt x } \right)}} + \dfrac{{x + 1}}{{\sqrt x }} - \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + \left( {x + 1} \right)\left( {\sqrt x - 1} \right) - \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + x\sqrt x - x + \sqrt x - 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + 25}}{{\sqrt x }}\\
b)Dkxd:x > 0;x\# 1\\
Q = \dfrac{{x + 25}}{{\sqrt x }}\\
= \sqrt x + \dfrac{{25}}{{\sqrt x }} \ge 2.\sqrt {\sqrt x .\dfrac{{25}}{{\sqrt x }}} = 10\\
\Leftrightarrow Q \ge 10\\
\Leftrightarrow GTNN:Q = 10\\
Khi:\sqrt x = \dfrac{{25}}{{\sqrt x }} \Leftrightarrow x = 25\left( {tmdk} \right)\\
Vậy\,x = 25
\end{array}$
