ĐKXĐ: x > 0; x ≠ 9

\(A = \frac{{x + 7}}{{\sqrt x + 3}} = \sqrt x - 3 + \frac{{16}}{{\sqrt x + 3}} = \sqrt x + 3 + \frac{{16}}{{\sqrt x + 3}} - 6\)

Áp dụng BĐT Cô-si cho 2 số không âm \(\sqrt x + 3\) và \(\frac{{16}}{{\sqrt x + 3}}\) ta được:

\(\sqrt x + 3 + \frac{{16}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\frac{{16}}{{\sqrt x + 3}}} = 2.4 = 8\)

\( \Rightarrow \sqrt x + 3 + \frac{{16}}{{\sqrt x + 3}} - 6 \ge 2 \Rightarrow A \ge 2\)

Dấu “=” xảy ra \( \Leftrightarrow \sqrt x + 3 = \frac{{16}}{{\sqrt x + 3}} \Leftrightarrow {\left( {\sqrt x + 3} \right)^2} = 16\)

\( \Leftrightarrow \left| {\sqrt x + 3} \right| = 4 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sqrt x + 3 = 4}\\{\sqrt x + 3 = - 4}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sqrt x = 1}\\{\sqrt x = - 7(L)}\end{array}} \right. \Leftrightarrow x = 1(TM)\)

Vậy \({A_{\min }} = 2\) khi x = 1.