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