Đáp án:
\(6 \ge m > 0;m \ne \dfrac{3}{{13}}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{m}{6}\\
\to \dfrac{1}{{\sqrt x + 1}} = \dfrac{m}{6}\\
\to 6 = m\sqrt x + m\\
\to \sqrt x = \dfrac{{6 - m}}{m}
\end{array}\)
Để \(P = \dfrac{m}{6}\) có nghiệm
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{6 - m}}{m} \ge 0\\
\dfrac{{6 - m}}{m} \ne 25\\
m \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
6 - m \ge 0\\
m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
6 - m \le 0\\
m < 0
\end{array} \right.
\end{array} \right.\\
6 - m \ne 25m\\
m \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
6 \ge m > 0\\
\left\{ \begin{array}{l}
m \ge 6\\
m < 0
\end{array} \right.\left( l \right)
\end{array} \right.\\
m \ne \left\{ {0;\dfrac{3}{{13}}} \right\}
\end{array} \right.\\
\to 6 \ge m > 0;m \ne \dfrac{3}{{13}}
\end{array}\)
