Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x = 16 \Rightarrow A = \dfrac{{2\sqrt {16} + 1}}{{16 + \sqrt {16} + 1}} = \dfrac{{2.4 + 1}}{{16 + 4 + 1}} = \dfrac{9}{{21}} = \dfrac{3}{7}\\
b,\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{\sqrt x }}{{1 - x}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - 1} \right)\\
= \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - 1} \right)\\
= \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\dfrac{{\sqrt x - \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x - 1} \right)\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
3,\\
M = \dfrac{P}{A} = \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}:\dfrac{{2\sqrt x + 1}}{{x + \sqrt x + 1}} = \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}.\dfrac{{x + \sqrt x + 1}}{{2\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{{\left( {x + \sqrt x } \right) + 1}}{{\sqrt x + 1}} = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + 1}}{{\sqrt x + 1}}\\
= \sqrt x + \dfrac{1}{{\sqrt x + 1}} = \left( {\sqrt x + 1} \right) + \dfrac{1}{{\sqrt x + 1}} - 1\\
\ge 2.\sqrt {\left( {\sqrt x + 1} \right).\dfrac{1}{{\sqrt x + 1}}} - 1 = 2 - 1 = 1\\
\Rightarrow {M_{\min }} = 1 \Leftrightarrow \sqrt x + 1 = \dfrac{1}{{\sqrt x + 1}} \Leftrightarrow \sqrt x + 1 = 1 \Leftrightarrow x = 0
\end{array}\)
