Đáp án:
\[{A_{\max }} = \dfrac{4}{{27}} \Leftrightarrow x = \dfrac{5}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{1}{{x - \sqrt {5x} + 8}}\,\,\,\,\,\left( {x \ge 0} \right)\\
M = x - \sqrt {5x} + 8\\
\Leftrightarrow 5M = 5\left( {x - \sqrt {5x} + 8} \right)\\
\Leftrightarrow 5M = 5x - 5\sqrt {5x} + 40\\
\Leftrightarrow 5M = {\sqrt {5x} ^2} - 2.\sqrt {5x} .\dfrac{5}{2} + \dfrac{{25}}{4} + \dfrac{{135}}{4}\\
\Leftrightarrow 5M = \left( {{{\sqrt {5x} }^2} - 2.\sqrt {5x} .\dfrac{5}{2} + {{\left( {\dfrac{5}{2}} \right)}^2}} \right) + \dfrac{{135}}{4}\\
\Leftrightarrow 5M = {\left( {\sqrt {5x} - \dfrac{5}{2}} \right)^2} + \dfrac{{135}}{4} \ge \dfrac{{135}}{4},\,\,\,\,\forall x \ge 0\\
\Rightarrow M \ge \dfrac{{27}}{4},\,\,\,\,\forall x \ge 0\\
\Rightarrow {M_{\min }} = \dfrac{{27}}{4} \Leftrightarrow {\left( {\sqrt {5x} - \dfrac{5}{2}} \right)^2} = 0 \Leftrightarrow \sqrt {5x} = \dfrac{5}{2} \Leftrightarrow 5x = \dfrac{{25}}{4} \Leftrightarrow x = \dfrac{5}{4}\\
\Rightarrow A = \dfrac{1}{M} \le \dfrac{1}{{\dfrac{{27}}{4}}} = \dfrac{4}{{27}},\,\,\,\forall x \ge 0\\
\Rightarrow {A_{\max }} = \dfrac{4}{{27}} \Leftrightarrow x = \dfrac{5}{4}
\end{array}\)
Vậy \({A_{\max }} = \dfrac{4}{{27}} \Leftrightarrow x = \dfrac{5}{4}\)
