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