Đáp án:
$\begin{array}{l}
M = \dfrac{{{x^2} + 2x - 9}}{{x - 3}}\\
= \dfrac{{{x^2} - 3x + 5x - 15 + 6}}{{x - 3}}\\
= \dfrac{{x\left( {x - 3} \right) + 5\left( {x - 3} \right) + 6}}{{x - 3}}\\
= \dfrac{{\left( {x + 5} \right)\left( {x - 3} \right) + 6}}{{x - 3}}\\
= x + 5 + \dfrac{6}{{x - 3}}\\
= x - 3 + \dfrac{6}{{x - 3}} + 8\\
Do:x > 3 \Rightarrow x - 3 > 0\\
Theo\,Cô - si:\\
x - 3 + \dfrac{6}{{x - 3}} \ge 2\sqrt {\left( {x - 3} \right).\dfrac{6}{{x - 3}}} = 2\sqrt 6 \\
\Rightarrow M \ge 2\sqrt 6 + 8\\
\Rightarrow GTNN:M = 2\sqrt 6 + 8\\
Khi:x - 3 = \dfrac{6}{{x - 3}}\\
\Rightarrow {\left( {x - 3} \right)^2} = 6\\
\Rightarrow x - 3 = \sqrt 6 \\
\Rightarrow x = \sqrt 6 + 3
\end{array}$
