2 câu trả lời
Đáp án:
\[{M_{\max }} = 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
M = \dfrac{{6\sqrt x }}{{x + \sqrt x + 1}}\,\,\,\,\,\left( {x \ge 0} \right)\\
M - 2 = \dfrac{{6\sqrt x }}{{x + \sqrt x + 1}} - 2\\
= \dfrac{{6\sqrt x - 2.\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{6\sqrt x - 2x - 2\sqrt x - 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{ - 2x + 4\sqrt x - 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{ - 2.\left( {x - 2\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{ - 2.{{\left( {\sqrt x - 1} \right)}^2}}}{{x + \sqrt x + 1}}\\
{\left( {\sqrt x - 1} \right)^2} \ge 0,\,\,\,\forall x \ge 0\\
\Rightarrow - 2.{\left( {\sqrt x - 1} \right)^2} \le 0,\,\,\,\forall x \ge 0\\
x + \sqrt x + 1 \ge 1 > 0,\,\,\,\,\forall x \ge 0\\
\Rightarrow M - 2 = \dfrac{{ - 2{{\left( {\sqrt x - 1} \right)}^2}}}{{x + \sqrt x + 1}} \le 0,\,\,\,\forall x \ge 0\\
\Leftrightarrow M \le 2,\,\,\,\forall x \ge 0\\
\Rightarrow {M_{\max }} = 2 \Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 0 \Leftrightarrow \sqrt x - 1 = 0 \Leftrightarrow x = 1\\
\Rightarrow {M_{\max }} = 2
\end{array}\)
Đáp án:
Giải thích các bước giải:
$M = \dfrac{{6\sqrt x }}{{x + \sqrt x + 1}}\\ĐK:x \ge 0\\ M - 2 = \dfrac{{6\sqrt x }}{{x + \sqrt x + 1}} - 2 = \dfrac{{6\sqrt x - 2.\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\ = \dfrac{{6\sqrt x - 2x - 2\sqrt x - 2}}{{x + \sqrt x + 1}} = \dfrac{{ - 2x + 4\sqrt x - 2}}{{x + \sqrt x + 1}}\\ = \dfrac{{ - 2.\left( {x - 2\sqrt x + 1} \right)}}{{x + \sqrt x + 1}} = \dfrac{{ - 2.{{\left( {\sqrt x - 1} \right)}^2}}}{{x + \sqrt x + 1}}\\ {\left( {\sqrt x - 1} \right)^2} \ge 0\forall x \ge 0\\ \Rightarrow - 2.{\left( {\sqrt x - 1} \right)^2} \le 0\forall x \ge 0 x + \sqrt x + 1 \ge 1 > 0,\forall x \ge 0\\ \Rightarrow M - 2 = \dfrac{{ - 2{{\left( {\sqrt x - 1} \right)}^2}}}{{x + \sqrt x + 1}} \le 0\forall x \ge 0\ \Leftrightarrow M \le 2\forall x \ge 0\\ \Rightarrow {M_{\max }} = 2 \Leftrightarrow x = 1$