Đáp án:
\(\min A = 0 \Leftrightarrow x = 0\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}A = \dfrac{{5\sqrt x }}{{\sqrt x + 1}}\,\,\left( {x \ge 0} \right)\\A = \dfrac{{5\sqrt x + 5 - 5}}{{\sqrt x + 1}}\\A = \dfrac{{5\left( {\sqrt x + 1} \right) - 5}}{{\sqrt x + 1}}\\A = 5 - \dfrac{5}{{\sqrt x + 1}}\end{array}\)
Vì \(\sqrt x \ge 0\,\,\forall x \ge 0\) nên \(\sqrt x + 1 \ge 1\,\,\forall x \ge 0\)
\(\begin{array}{l} \Rightarrow \dfrac{5}{{\sqrt x + 1}} \le 5\,\,\forall x \ge 0 \Leftrightarrow - \dfrac{5}{{\sqrt x + 1}} \ge - 5\,\,\forall x \ge 0\\ \Rightarrow 5 - \dfrac{5}{{\sqrt x + 1}} \ge 0\,\,\forall x \ge 0\\ \Rightarrow A \ge 0\,\,\,\forall x \ge 0\\ \Rightarrow \min A = 0 \Leftrightarrow x = 0\end{array}\)
