Đáp án:
\(3 \ge m > 0\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{x - \sqrt x - 4\sqrt x + 6}}{{\sqrt x \left( {\sqrt x - 3} \right)}} = \dfrac{{x - 5\sqrt x + 6}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 3} \right)}} = \dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
P = \dfrac{B}{A} = \dfrac{{\sqrt x - 2}}{{\sqrt x }}:\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
P + m = 1\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} + m = 1\\
\to \dfrac{{\sqrt x - 2 + m\sqrt x + m - \sqrt x - 1}}{{\sqrt x + 1}} = 0\\
\to m\sqrt x + m - 3 = 0\\
\to \sqrt x = \dfrac{{3 - m}}{m}
\end{array}\)
Theo yêu cầu bài toán
\(\begin{array}{l}
\Leftrightarrow \dfrac{{3 - m}}{m} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - m \ge 0\\
m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - m \le 0\\
m < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
3 \ge m > 0\\
\left\{ \begin{array}{l}
m \ge 3\\
m < 0
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 3 \ge m > 0
\end{array}\)
