Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
A = \left( {\frac{{3\sqrt x }}{{x - 4}} + \frac{1}{{\sqrt x + 2}}} \right):\frac{2}{{\sqrt x - 2}}\\
= \left( {\frac{{3\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\frac{2}{{\sqrt x - 2}}\\
= \frac{{4\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x - 2}}{2}\\
= \frac{{2\sqrt x - 1}}{{\sqrt x + 2}}
\end{array}\)
b,
\[\begin{array}{l}
A\left( {\sqrt x + 2} \right) = x\\
\Leftrightarrow \frac{{2\sqrt x - 1}}{{\sqrt x + 2}}.\left( {\sqrt x + 2} \right) = x\\
\Leftrightarrow 2\sqrt x - 1 = x\\
\Leftrightarrow x - 2\sqrt x + 1 = 0\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 0\\
\Leftrightarrow x = 1
\end{array}\]
c,
Ta có:
\[\begin{array}{l}
A\left( {\sqrt x + 2} \right) = 2\sqrt x - 1\\
\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
2\sqrt x - 1 \ge - 1\\
2\sqrt x - 1 \ne 3
\end{array} \right.
\end{array}\]
Do đó, để phương trình đã cho có nghiệm thì \(\left\{ \begin{array}{l}
m \ge - 1\\
m \ne 3
\end{array} \right.\)
