2 câu trả lời
$y = \dfrac{3 sinx + 2 cosx}{2 + sinx + cosx}$
$y' = \dfrac{(3 cosx -2 sinx)(2 + sinx + cosx)-(3 sinx + 2 cosx)(cosx - sinx)}{(2 + sinx + cosx)^2}$
$y' = \dfrac{6 cosx + 3 sinx cosx + 3 cos^2x - 4 sinx - 2 sin^2x - 2 sinx cosx -(3 sinx cosx - 3 sin^2x + 2 cos^2x - 2 sinx cosx}{(2 + sinx + cosx)^2}$
$y' = \dfrac{1 + 6cosx - 4 sinx}{(2 + sinx + cosx)^2}$
y'=0 <-> 1 + 6cosx - 4 sinx = 0 hay 4 sinx - 6 cosx = 1.
Chia ca 2 ve cho \sqrt{52}, dat $cos \alpha=4/\sqrt{52}, sin \alpha= 6/\sqrt{52}, sin \beta=1/\sqrt{52}$. Khi do
$sin x cos \alpha - cosx sin \alpha = sin \beta$ hay $sin(x+\alpha) = sin \beta$ vay $x_1 = \beta - \alpha + 2 k \pi$ hoac $x_2 = \pi - \alpha - \beta + 2k \pi$.
Voi $x_1 = \beta - \alpha + 2k \pi$ va ap dung cong thuc
cos(a$\pm$b) = cosa cosb $\mp$ sina sinb va sin(a$\pm$b) = sina cosb $\pm$ sinb cosa thay vao ptrinh ta co
$y(x_1) =\dfrac{12-5\sqrt{51}}{61 - \sqrt{51}}$.
$y(x_2) = \dfrac{13}{5 - \sqrt{51}}$.
$\begin{array}{l}y = \dfrac{1 - 3\sin x + 2\cos x}{2 + \sin x + \cos x}\\ \Leftrightarrow 2y + y\sin x + y\cos x = 1 - 3\sin x + 2\cos x\\ \Leftrightarrow (y + 3)\sin x + (y - 2)\cos x = 1 - 2y\\ \text{Phương trình có nghiệm}\,\,\Leftrightarrow (y + 3)^2 + (y - 2)^2 \geq (1 - 2y)^2\\ \Leftrightarrow y^2 - 3y - 6 \leq 0\\ \Leftrightarrow \dfrac{3 - \sqrt{33}}{2} \leq y \leq \dfrac{3 + \sqrt{33}}{2}\\ \Rightarrow \begin{cases}\min y = \dfrac{3 - \sqrt{33}}{2}\\\max y = \dfrac{3 + \sqrt{33}}{2}\end{cases}\end{array}$
