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