Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
x = 9 \Rightarrow A = \dfrac{{3\sqrt x }}{{\sqrt x + 1}} = \dfrac{{3.\sqrt 9 }}{{\sqrt 9 + 1}} = \dfrac{{3.3}}{{3 + 1}} = \dfrac{9}{4}\\
2,\\
B = \dfrac{{2x + \sqrt x - 4}}{{x - \sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} + \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{2x + \sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} + \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {2x + \sqrt x - 4} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 1}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {2x + \sqrt x - 4} \right) - \left( {x - 4} \right) + \sqrt x + 1}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}}\\
P = A.B = \dfrac{{3\sqrt x }}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} = \dfrac{{3\sqrt x }}{{\sqrt x - 2}}\\
3,\\
P \ge 2 \Rightarrow \dfrac{{3\sqrt x }}{{\sqrt x - 2}} \ge 2\\
\Leftrightarrow \dfrac{{3\sqrt x }}{{\sqrt x - 2}} - 2 \ge 0\\
\Leftrightarrow \dfrac{{3\sqrt x - 2\left( {\sqrt x - 2} \right)}}{{\sqrt x - 2}} \ge 0\\
\Leftrightarrow \dfrac{{\sqrt x + 4}}{{\sqrt x - 2}} \ge 0\\
\Leftrightarrow \sqrt x - 2 > 0\\
\Leftrightarrow x > 4
\end{array}\)