Đáp án:
Min=2
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{x + 3\sqrt x - 2 - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
c)DK:x \ge 0\\
P = \dfrac{A}{B} = \dfrac{{x + 3}}{{\sqrt x + 3}}:\dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{x + 3}}{{\sqrt x + 1}} = \dfrac{{x - 1 + 4}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 4}}{{\sqrt x + 1}}\\
= \left( {\sqrt x - 1} \right) + \dfrac{4}{{\sqrt x + 1}}\\
= \left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} - 2\\
Do:x \ge 0\\
BDT:Co - si:\\
\left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{4}{{\sqrt x + 1}}} = 4\\
\to \left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} - 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x + 1} \right) = \dfrac{4}{{\sqrt x + 1}}\\
\to {\left( {\sqrt x + 1} \right)^2} = 4\\
\to \sqrt x + 1 = 2\\
\to x = 1
\end{array}\)
