Đáp án:
a) \(\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 2}} - \dfrac{{3\sqrt x }}{{x + \sqrt x - 2}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x - 1 - 3\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x - 2\sqrt x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
b)S = A.B = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}.\dfrac{{\sqrt x + 3}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 1}}{{\sqrt x + 2}} = 1 + \dfrac{1}{{\sqrt x + 2}}\\
Do:\sqrt x \ge 0\forall x \ge 0;\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{1}{{\sqrt x + 2}} \le \dfrac{1}{2}\\
\to 1 + \dfrac{1}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to MaxS = \dfrac{3}{2}\\
\Leftrightarrow x = 0
\end{array}\)
