Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 9\\
b.P = \frac{{\sqrt x \left( {3 - \sqrt x } \right) + 2x}}{{9 - x}}:\left( {\frac{{\sqrt x - 1 - 2\left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right)\\
= \frac{{3\sqrt x + x}}{{9 - x}}.\frac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{5 - \sqrt x }} = \frac{{ - 3x - x\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {5 - \sqrt x } \right)}}\\
P = \frac{{ - 4}}{3} = \frac{{ - 3x - x\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {5 - \sqrt x } \right)}}\\
\to (4\sqrt x + 12)\left( {5 - \sqrt x } \right) = 9x + 3x\sqrt x \\
\to 20\sqrt x - 4x + 60 - 12\sqrt x = 9x + 3x\sqrt x \\
\to 13x + 3x\sqrt x - 8\sqrt x - 60 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = - 3\left( l \right)\\
\sqrt x = \frac{{ - 10}}{3}\left( l \right)
\end{array} \right. \to x = 4\left( {TM} \right)
\end{array}\)
