Đáp án:
\(Max = \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
C = \dfrac{x}{{\sqrt x - 3}}\\
N = \dfrac{C}{{\sqrt x - 1 + C}} = \dfrac{{\dfrac{x}{{\sqrt x - 3}}}}{{\sqrt x - 1 + \dfrac{x}{{\sqrt x - 3}}}}\\
= \dfrac{x}{{\sqrt x - 3}}:\left( {\dfrac{{x - 3\sqrt x - \sqrt x + 3 + x}}{{\sqrt x - 3}}} \right)\\
= \dfrac{x}{{\sqrt x - 3}}.\dfrac{{\sqrt x - 3}}{{2x - 4\sqrt x + 3}}\\
= \dfrac{x}{{2x - 4\sqrt x + 3}}\\
\dfrac{1}{N} = \dfrac{{2x - 4\sqrt x + 3}}{x}\\
= 2 - \dfrac{4}{{\sqrt x }} + \dfrac{3}{x}\\
= \dfrac{3}{x} - 2.\dfrac{{\sqrt 3 }}{{\sqrt x }}.\dfrac{2}{{\sqrt 3 }} + \dfrac{4}{3} + \dfrac{2}{3}\\
= {\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{2}{3}\\
\to N = \dfrac{1}{{{{\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }}} \right)}^2} + \dfrac{2}{3}}}\\
Do:{\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }}} \right)^2} \ge 0\forall x > 0\\
\to {\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{2}{3} \ge \dfrac{2}{3}\\
\to \dfrac{1}{{{{\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }}} \right)}^2} + \dfrac{2}{3}}} \le 1:\dfrac{2}{3} = \dfrac{3}{2}\\
\to Max = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{{\sqrt x }} - \dfrac{2}{{\sqrt 3 }} = 0\\
\Leftrightarrow \sqrt x = \dfrac{3}{2}\\
\to x = \dfrac{9}{4}
\end{array}\)
