Giải thích các bước giải:
\(\begin{array}{l}
b,\\
x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{4}{{4 + 2\sqrt 3 }} = \dfrac{{{2^2}}}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
\Rightarrow x = \dfrac{2}{{\sqrt 3 + 1}} = \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} = \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} = \sqrt 3 - 1\\
M = \dfrac{{2\sqrt x }}{{x + 1}} = \dfrac{{2.\left( {\sqrt 3 - 1} \right)}}{{\dfrac{2}{{2 + \sqrt 3 }} + 1}} = \dfrac{{2\sqrt 3 - 2}}{{\dfrac{{2 + 2 + \sqrt 3 }}{{2 + \sqrt 3 }}}} = \dfrac{{2\sqrt 3 - 2}}{{\dfrac{{4 + \sqrt 3 }}{{2 + \sqrt 3 }}}}\\
= \dfrac{{\left( {2\sqrt 3 - 2} \right)\left( {2 + \sqrt 3 } \right)}}{{4 + \sqrt 3 }} = \dfrac{{4\sqrt 3 + 6 - 4 - 2\sqrt 3 }}{{4 + \sqrt 3 }} = \dfrac{{2\sqrt 3 + 2}}{{4 + \sqrt 3 }}\\
= \dfrac{{\left( {2\sqrt 3 + 2} \right)\left( {4 - \sqrt 3 } \right)}}{{\left( {4 - \sqrt 3 } \right)\left( {4 + \sqrt 3 } \right)}} = \dfrac{{8\sqrt 3 - 6 + 8 - 2\sqrt 3 }}{{16 - 3}} = \dfrac{{6\sqrt 3 + 2}}{{13}}\\
c,\\
M = \dfrac{{2\sqrt x }}{{x + 1}}\\
M - 1 = \dfrac{{2\sqrt x }}{{x + 1}} - 1 = \dfrac{{2\sqrt x - x - 1}}{{x + 1}} = \dfrac{{ - \left( {x - 2\sqrt x + 1} \right)}}{{x + 1}} = - \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{x + 1}} \le 0,\,\,\,\forall x \ge 0\\
\Rightarrow M - 1 \le 0\\
\Leftrightarrow M \le 1,\,\,\,\forall x \ge 0\\
\Rightarrow {M_{\max }} = 1 \Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 0 \Leftrightarrow x = 1
\end{array}\)
