\(y=\dfrac{sinx+2cosx+1}{sinx+cosx+2}\)
Thấy : \(sinx+cosx+2\ge-1-1+2=0\) . " = " ko xảy ra nên : \(sinx+cosx+2>0\)
Suy ra : \(\left(y-1\right)sinx+\left(y-2\right)cosx=1-2y\) (*)
(*) có no \(\Leftrightarrow\left(y-1\right)^2+\left(y-2\right)^2\ge\left(1-2y\right)^2\Leftrightarrow2y^2-6y+5\ge4y^2-4y+1\Leftrightarrow-2y^2-2y+4\ge0\)
\(\Leftrightarrow-y^2-y+2\ge0\) \(\Leftrightarrow-2\le y\le1\)
Suy ra : Max y = 1 . Chọn B
21 : \(cosx-\sqrt{3}sinx=0\)
cos x = 0 thay vào : sin x = 0 ( L )
cos x khác 0 \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\); ta có : \(1-\sqrt{3}tanx=0\Leftrightarrow tanx=\dfrac{1}{\sqrt{3}}\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\left(k\in Z\right)\)