Toán Tìm GTLN,GTNN của hàm số: y= $\frac{2sinx-1}{cosx+2}$ 05/07/2021 By Adalynn Tìm GTLN,GTNN của hàm số: y= $\frac{2sinx-1}{cosx+2}$
Đáp án: $\begin{array}{l}y = \dfrac{{2\sin x – 1}}{{\cos x + 2}}\\ \Rightarrow y.\cos x + 2y = 2\sin x – 1\\ \Rightarrow y.\cos x – 2\sin x = – 2y – 1\\Dk:{y^2} + {\left( { – 2} \right)^2} \ge {\left( { – 2y – 1} \right)^2}\\ \Rightarrow {y^2} + 4 \ge 4{y^2} + 4y + 1\\ \Rightarrow 3{y^2} + 4y – 3 \le 0\\ \Rightarrow \dfrac{{ – \sqrt {13} – 2}}{3} \le y \le \dfrac{{\sqrt {13} – 2}}{3}\\ \Rightarrow \left\{ \begin{array}{l}GTNN:y = \dfrac{{ – \sqrt {13} – 2}}{3}\\GTLN:y = \dfrac{{\sqrt {13} – 2}}{3}\end{array} \right.\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
y = \dfrac{{2\sin x – 1}}{{\cos x + 2}}\\
\Rightarrow y.\cos x + 2y = 2\sin x – 1\\
\Rightarrow y.\cos x – 2\sin x = – 2y – 1\\
Dk:{y^2} + {\left( { – 2} \right)^2} \ge {\left( { – 2y – 1} \right)^2}\\
\Rightarrow {y^2} + 4 \ge 4{y^2} + 4y + 1\\
\Rightarrow 3{y^2} + 4y – 3 \le 0\\
\Rightarrow \dfrac{{ – \sqrt {13} – 2}}{3} \le y \le \dfrac{{\sqrt {13} – 2}}{3}\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = \dfrac{{ – \sqrt {13} – 2}}{3}\\
GTLN:y = \dfrac{{\sqrt {13} – 2}}{3}
\end{array} \right.
\end{array}$