Đáp án:
m>0
Giải thích các bước giải:
\(\begin{array}{l}
B = \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 4} \right) + 5\sqrt x + 12}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \frac{{x - \sqrt x - 12 + 5\sqrt x + 12}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \frac{{x + 4\sqrt x }}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \frac{{\sqrt x }}{{\sqrt x - 4}}\\
Có:\frac{A}{B} = m + 1\\
\to \frac{{\sqrt x + 3}}{{\sqrt x - 4}}:\frac{{\sqrt x }}{{\sqrt x - 4}} = m + 1\\
\to \frac{{\sqrt x + 3}}{{\sqrt x }} = m + 1\\
\to \sqrt x + 3 = m\sqrt x + \sqrt x \\
\to m\sqrt x = 3\\
\to \sqrt x = \frac{3}{m}\\
\to x = \frac{9}{{{m^2}}}
\end{array}\)
Để \(\frac{A}{B}\) có nghiệm
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
\frac{3}{m} \ge 0
\end{array} \right.\\
\to m > 0
\end{array}\)
