Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
B = x + 4\sqrt x \\
DKXD:\,\,\,\,x \ge 0\\
x \ge 0 \Rightarrow x + 4\sqrt x \ge 0\\
\Rightarrow {B_{\min }} = 0 \Leftrightarrow x = 0\\
4,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x - 1 \ne 0\\
\sqrt x + 1 \ne 0\\
x - 1 \ne 0\\
4\sqrt x - 8 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 4
\end{array} \right.\\
A = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x }}{{x - 1}}} \right):\dfrac{{4\sqrt x - 8}}{{1 - x}}\\
= \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\dfrac{{4\sqrt x - 8}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} - 8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{4\sqrt x - 8}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{\left( {x + 2\sqrt x + 1} \right) - \left( {x - 2\sqrt x + 1} \right) - 8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{4\sqrt x - 8}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{ - 4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {1 - \sqrt x } \right).\left( {1 + \sqrt x } \right)}}{{4.\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}
\end{array}\)
