Đáp án:
Min=2
Giải thích các bước giải:
\(\begin{array}{l}
3)M = \dfrac{{x + 7}}{{\sqrt x + 3}} = \dfrac{{x - 9 + 16}}{{\sqrt x + 3}} = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 16}}{{\sqrt x + 3}}\\
= \left( {\sqrt x - 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} = \left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} - 6\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{16}}{{\sqrt x + 3}}} = 8\\
\to \left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} - 6 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x + 3} \right) = \dfrac{{16}}{{\sqrt x + 3}}\\
\to {\left( {\sqrt x + 3} \right)^2} = 16\\
\to \sqrt x + 3 = 4\\
\to x = 1
\end{array}\)
