Tìm min max của hs:
$y = \frac{3\cos^{4} x + 4\sin^{2} x}{3\sin^{4}x+2\cos^{2}x}$
Ta có \(y = \frac{{3{{\left( {1 - {{\sin }^2}x} \right)}^2} + 4{{\sin }^2}x}}{{3{{\sin }^4}x + 2\left( {1 - {{\sin }^2}x} \right)}}\)
Đặt \(t = {\sin ^2}x;0 \le t \le 1\). Khi đó \(y = \frac{{3{t^2} - 2t + 3}}{{3{t^2} - 2t + 2}} = 1 + \frac{1}{{3{t^2} - 2t + 2}}\)
Xét hàm số
    \(f\left( t \right) = 3{t^2} - 2t + 2   (0 \le t \le 1)\)
\(\Rightarrow f'\left( t \right) = 6t - 2 = 0 \Leftrightarrow t = \frac{1}{3}\)
Ta có bảng biến thiên như hình vẽ:


Suy ra \(\frac{5}{3} \le f\left( t \right) \le 3 \Leftrightarrow \frac{8}{5} \ge y \ge \frac{4}{3}\)
Vậy \(\min y  = \frac{4}{3}\)khi \({\sin ^2}x = 1\), chẳng hạn khi \(x = \frac{\pi }{2}\)
\(\max y  = \frac{8}{5}\)khi \({\sin ^2}x = \frac{1}{3}\)
Viết lại $y-1=\frac{3(\cos^4 x-\sin^4 x)+4\sin^2 x-2]cos^2 x}{3\sin^4 x+2\cos^2 x}$
$\Leftrightarrow y-1=\frac{3(\cos^2 x-\sin^2 x+4\sin^2 x-2\cos^2 x}{3\sin^4 x+2\cos^2 x}$
$\Leftrightarrow y-1=\frac{1}{3\sin^4 x+2\cos^2 x}$. Đặt $\sin^2 x=t, t \in [0;1]$, hàm số trở thành
  $y-1=\frac{1}{3t^2-2t+2}$. Gọi $f(t)=3t^2-2t+2$
Thấy rằng $\begin{cases}f(t)>0, \forall t \in [0;1],  do  \Delta'=-5<0  và  a=3>0 \\ -\frac{b}{2a}=\frac{1}{3} \in [0;1] \end{cases}$, suy ra:
* $\mathop {\min}\limits_{D} f(t)=-\frac{\Delta'}{a}=\frac{5}{3} \Rightarrow \mathop {\max}\limits_{D} (y-1)=\frac{3}{5} \Leftrightarrow \mathop {\max}\limits_{D} y=\frac{3}{5}+1=\frac{8}{5}$, có được khi và chỉ khi $t=\frac{1}{3} \Leftrightarrow \sin^2 x=\frac{1}{3}$
* $\mathop {\max}\limits_{D} f(t)=\max\left\{ {f(0);f(1)} \right\}=\max \left\{ {2;3} \right\}=f(1)=3 \Rightarrow \mathop {\min}\limits_{D} (y-1)=\frac{1}{3}$
$\Leftrightarrow \mathop {\min}\limits_{D} y=\frac{1}{3}+1=\frac{4}{3}$, có được khi và chỉ khi $t=1 \Leftrightarrow \sin^2 x=1$
Tóm lại: GTLN của hàm số bằng $\frac{8}{5}$; GTNN của hàm số bằng $\frac{4}{3}$

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