Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
A = \frac{{x\sqrt x - 3}}{{x - 2\sqrt x - 3}} - \frac{{2\left( {\sqrt x - 3} \right)}}{{\sqrt x + 1}} + \frac{{\sqrt x + 3}}{{3 - \sqrt x }}\\
= \frac{{x\sqrt x - 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}} - \frac{{2{{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{x\sqrt x - 3 - 2.\left( {x - 6\sqrt x + 9} \right) - \left( {x + 4\sqrt x + 3} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{x\sqrt x - 3 - 2x + 12\sqrt x - 18 - x - 4\sqrt x - 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{x\sqrt x - 3x + 8\sqrt x - 24}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\left( {\sqrt x - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x + 1} \right).\left( {\sqrt x - 3} \right)}}\\
= \frac{{x + 8}}{{\sqrt x + 1}}\\
P = \frac{{x + 8}}{{\sqrt x + 1}} = \frac{{\left( {x - 1} \right) + 9}}{{\sqrt x + 1}} = \left( {\sqrt x - 1} \right) + \frac{9}{{\sqrt x + 1}}\\
= \left( {\sqrt x + 1} \right) + \frac{9}{{\sqrt x + 1}} - 2 \ge 2.\sqrt {\left( {\sqrt x + 1} \right).\frac{9}{{\sqrt x + 1}}} - 2 = 2.3 - 2 = 4\\
\Rightarrow {P_{\min }} = 4 \Leftrightarrow x = 4
\end{array}\)
