\(y=\dfrac{2\left(2cos^2x-1\right)+3sin2x+3}{3-sin2x+cos2x}=\dfrac{3sin2x+2cos2x+3}{3-sin2x+cos2x}\left(\cdot\right)\)\(\left(\cdot\right)\Leftrightarrow y\left(3-sin2x+cos2x\right)=3sin2x+2cos2x+3\) \(\Leftrightarrow\left(y+3\right)sin2x+\left(2-y\right)cos2x=3y-3\left(1\right)\)Để phương trình (1) có nghiệm thì:\(\left(y+3\right)^2+\left(2-y\right)^2\ge\left(3y-3\right)^2\)\(\Leftrightarrow7y^2-20y-4\le0\)\(\Leftrightarrow\dfrac{10-8\sqrt{2}}{7}\le y\le\dfrac{10+8\sqrt{2}}{7}\)=> min, max của y Câu trả lời: \(y=\dfrac{2\left(2cos^2x-1\right)+3sin2x+3}{3-sin2x+cos2x}=\dfrac{3sin2x+2cos2x+3}{3-sin2x+cos2x}\left(\cdot\right)\)\(\left(\cdot\right)\Leftrightarrow y\left(3-sin2x+cos2x\right)=3sin2x+2cos2x+3\) \(\Leftrightarrow\left(y+3\right)sin2x+\left(2-y\right)cos2x=3y-3\left(1\right)\)Để phương trình (1) có nghiệm thì:\(\left(y+3\right)^2+\left(2-y\right)^2\ge\left(3y-3\right)^2\)\(\Leftrightarrow7y^2-20y-4\le0\)\(\Leftrightarrow\dfrac{10-8\sqrt{2}}{7}\le y\le\dfrac{10+8\sqrt{2}}{7}\)=> min, max của y