Đáp án:
Min=6
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{\sqrt x - 1}}{{x + 3}}\\
\dfrac{1}{P} = \dfrac{{x + 3}}{{\sqrt x - 1}} = \dfrac{{x - 1 + 4}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 4}}{{\sqrt x - 1}}\\
= \sqrt x - 1 + \dfrac{4}{{\sqrt x - 1}} + 2\\
Do:x > 1\\
\to \sqrt x - 1 + \dfrac{4}{{\sqrt x - 1}} \ge 2\sqrt {\left( {\sqrt x - 1} \right).\dfrac{4}{{\sqrt x - 1}}} \\
\to \sqrt x - 1 + \dfrac{4}{{\sqrt x - 1}} \ge 2.2\\
\to \sqrt x - 1 + \dfrac{4}{{\sqrt x - 1}} \ge 4\\
\to \sqrt x - 1 + \dfrac{4}{{\sqrt x - 1}} + 2 \ge 6\\
\to Min = 6\\
\Leftrightarrow \sqrt x - 1 = \dfrac{4}{{\sqrt x - 1}}\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 4\\
\to \left| {\sqrt x - 1} \right| = 2\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 2\\
\sqrt x - 1 = - 2\left( l \right)
\end{array} \right.\\
\to \sqrt x = 3\\
\to x = 9
\end{array}\)
