Bài tập 5.21 trang 203 SBT Toán 11

\(\begin{array}{l}
f\prime \left( x \right) = 1 - 2.\frac{{2x}}{{2\sqrt {{x^2} + 12} }} = 1 - \frac{{2x}}{{\sqrt {{x^2} + 12} }}\\
 \Rightarrow f\prime \left( x \right) \le 0 \Leftrightarrow 1 - \frac{{2x}}{{\sqrt {{x^2} + 12} }} \le 0\\
 \Leftrightarrow \frac{{\sqrt {{x^2} + 12}  - 2x}}{{\sqrt {{x^2} + 12} }} \le 0\\
 \Rightarrow \sqrt {{x^2} + 12}  - 2x \le 0\\
 \Leftrightarrow \sqrt {{x^2} + 12}  \le 2x\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} + 12 \le 4{x^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} \ge 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
x \le  - 2\\
x \ge 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 2
\end{array}\)

-- Mod Toán 11 HỌC247