Toán tìm gtnn gtln y=3sin2x+cos2x/sin2x+4cos^2x+1 27/07/2021 By Jade tìm gtnn gtln y=3sin2x+cos2x/sin2x+4cos^2x+1
$y = \dfrac{3\sin2x + \cos2x}{\sin2x + 4\cos^2x +1}$ $\Leftrightarrow y = \dfrac{3\sin2x + \cos2x}{\sin2x + 2(1 +\cos2x) +1}$ $\Leftrightarrow y = \dfrac{3\sin2x + \cos2x}{\sin2x + 2\cos2x +3}$ $\Leftrightarrow y\sin2x + 2y\cos2x + 3y = 3\sin2x + \cos2x $ $\Leftrightarrow (y -3)\sin2x + (2y -1)\cos2x + 3y = 0$ Phương trình có nghiệm $\Leftrightarrow (y -3)^2 + (2y -1)^2 \geq (3y)^2$ $\Leftrightarrow 2y^2 + 5y – 5 \leq 0$ $\Leftrightarrow \dfrac{-5 -\sqrt{65}}{4} \leq y \leq \dfrac{-5 +\sqrt{65}}{4}$ Vậy $\min y = \dfrac{-5 -\sqrt{65}}{4}$ $\max y = \dfrac{-5 +\sqrt{65}}{4}$ Trả lời
$y = \dfrac{3\sin2x + \cos2x}{\sin2x + 4\cos^2x +1}$
$\Leftrightarrow y = \dfrac{3\sin2x + \cos2x}{\sin2x + 2(1 +\cos2x) +1}$
$\Leftrightarrow y = \dfrac{3\sin2x + \cos2x}{\sin2x + 2\cos2x +3}$
$\Leftrightarrow y\sin2x + 2y\cos2x + 3y = 3\sin2x + \cos2x $
$\Leftrightarrow (y -3)\sin2x + (2y -1)\cos2x + 3y = 0$
Phương trình có nghiệm
$\Leftrightarrow (y -3)^2 + (2y -1)^2 \geq (3y)^2$
$\Leftrightarrow 2y^2 + 5y – 5 \leq 0$
$\Leftrightarrow \dfrac{-5 -\sqrt{65}}{4} \leq y \leq \dfrac{-5 +\sqrt{65}}{4}$
Vậy $\min y = \dfrac{-5 -\sqrt{65}}{4}$
$\max y = \dfrac{-5 +\sqrt{65}}{4}$