Đáp án:
\[{A_{\min }} = - \dfrac{1}{{20}} \Leftrightarrow x = 0\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge 0\)
Ta có:
\(\begin{array}{l}
A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 20}} = \dfrac{{\left( {\sqrt x + 20} \right) - 21}}{{\sqrt x + 20}} = 1 - \dfrac{{21}}{{\sqrt x + 20}}\\
\sqrt x \ge 0 \Rightarrow \sqrt x + 20 \ge 20\\
\Rightarrow \dfrac{{21}}{{\sqrt x + 20}} \le \dfrac{{21}}{{20}}\\
\Rightarrow 1 - \dfrac{{21}}{{\sqrt x + 20}} \ge 1 - \dfrac{{21}}{{20}} = - \dfrac{1}{{20}}\\
\Rightarrow A \ge - \dfrac{1}{{20}},\,\,\,\forall x \ge 0\\
\Rightarrow {A_{\min }} = - \dfrac{1}{{20}} \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0
\end{array}\)
Vậy \({A_{\min }} = - \dfrac{1}{{20}} \Leftrightarrow x = 0\)
