Đáp án:
\(Min = - \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
\dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}} = \dfrac{{2\left( {\sqrt x + 3} \right) - 7}}{{\sqrt x + 3}}\\
= 2 - \dfrac{7}{{\sqrt x + 3}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 3 \ge 3\\
\to \dfrac{7}{{\sqrt x + 3}} \le \dfrac{7}{3}\\
\to - \dfrac{7}{{\sqrt x + 3}} \ge - \dfrac{7}{3}\\
\to 2 - \dfrac{7}{{\sqrt x + 3}} \ge 2 - \dfrac{7}{3}\\
\to 2 - \dfrac{7}{{\sqrt x + 3}} \ge - \dfrac{1}{3}\\
\to Min = - \dfrac{1}{3}\\
\Leftrightarrow x = 0
\end{array}\)
