Đáp án:

 a. \(\dfrac{{4x}}{{\sqrt x  - 3}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x > 0;x \ne 4\\
P = \left[ {\dfrac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right]:\left[ {\dfrac{{\sqrt x  - 1 - 2\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}} \right]\\
 = \dfrac{{8\sqrt x  - 4x + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{ - \sqrt x \left( {2 - \sqrt x } \right)}}{{ - \sqrt x  + 3}}\\
 = \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\
 = \dfrac{{4\sqrt x .\sqrt x }}{{\sqrt x  - 3}}\\
 = \dfrac{{4x}}{{\sqrt x  - 3}}\\
b.P =  - 1\\
 \to \dfrac{{4x}}{{\sqrt x  - 3}} =  - 1\\
 \to 4x =  - \sqrt x  + 3\\
 \to 4x + \sqrt x  - 3 = 0\\
 \to \left[ \begin{array}{l}
\sqrt x  = \dfrac{3}{4}\\
\sqrt x  =  - 1\left( l \right)
\end{array} \right.\\
 \to x = \dfrac{9}{{16}}\\
c.m\left( {\sqrt x  - 3} \right).\dfrac{{4x}}{{\sqrt x  - 3}} > x + 1\\
 \to 4xm > x + 1\\
 \to x + 1 - 4xm < 0\\
 \to x\left( {1 - 4m} \right) + 1 < 0\\
 \to x <  - \dfrac{1}{{1 - 4m}}\\
Do:x > 9\\
 \to  - \dfrac{1}{{1 - 4m}} > 9\\
 \to \dfrac{1}{{1 - 4m}} <  - 9\\
 \to \dfrac{{1 + 9 - 36m}}{{1 - 4m}} < 0\\
 \to \dfrac{{10 - 36m}}{{1 - 4m}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
10 - 36m > 0\\
1 - 4m < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
10 - 36m < 0\\
1 - 4m > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{5}{{18}} > m\\
m > \dfrac{1}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{5}{{18}} < m\\
m < \dfrac{1}{4}
\end{array} \right.\left( l \right)
\end{array} \right.
\end{array}\)