Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
a,\\
A = \left( {\dfrac{x}{{x\sqrt x - 4\sqrt x }} - \dfrac{6}{{3\sqrt x - 6}} + \dfrac{1}{{\sqrt x + 2}}} \right):\left( {\sqrt x - 2 + \dfrac{{10 - x}}{{\sqrt x + 2}}} \right)\\
= \left( {\dfrac{x}{{\sqrt x .\left( {x - 4} \right)}} - \dfrac{2}{{\sqrt x - 2}} + \dfrac{1}{{\sqrt x + 2}}} \right):\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) + 10 - x}}{{\sqrt x + 2}}\\
= \left( {\dfrac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{2}{{\sqrt x - 2}} + \dfrac{1}{{\sqrt x + 2}}} \right):\dfrac{{\left( {x - 4} \right) + 10 - x}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x - 2.\left( {\sqrt x + 2} \right) + \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{6}{{\sqrt x + 2}}\\
= \dfrac{{ - 6}}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{6}\\
= \dfrac{{ - 1}}{{\sqrt x - 2}} = \dfrac{1}{{2 - \sqrt x }}\\
b,\\
A < 2 \Leftrightarrow \dfrac{1}{{2 - \sqrt x }} - 2 < 0\\
\Leftrightarrow \dfrac{{1 - 2.\left( {2 - \sqrt x } \right)}}{{2 - \sqrt x }} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 3}}{{2 - \sqrt x }} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 3}}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x > 2\\
\sqrt x < \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > 4\\
0 \le x < \dfrac{9}{4}
\end{array} \right.\\
\end{array}\)
