Question

giúp e với ajaa

Answer 1

e.

Vì $-1\leq \cos (x+\frac{\pi}{6})\leq 1$

$\Rightarrow -5\leq 3\cos (x+\frac{\pi}{6})-2\leq 1$

Vậy $y_{\min}=-5; y_{\max}=1$

f.

Vì $-1\leq \cos x\leq 1$

$\Rightarrow 0\leq |\cos x|\leq 1$

$\Rightarrow -10\leq 2|\cos x|-10\leq -8$

Vậy $y_{\min}=-10; y_{\max}=-8$

f.

Vì $-1\leq \sin x\leq 1\Rightarrow \sin ^2x\leq 1$
$\Rightarrow 0\leq \sqrt{1+3\sin ^2x}\leq \sqrt{1+3}=2$

$\Rightarrow 3\leq 7-2\sqrt{1+3\sin ^2x}\leq 7$

Vậy $y_{\min}=3; y_{\max}=7$

h.

$y=2-4\sin ^2x\cos ^2x=2-(2\sin x\cos x)^2=2-\sin ^22x$

Vì $-1\leq \sin 2x\leq 1\Rightarrow 0\leq \sin ^22x\leq 1$

$\Rightarrow 1\leq 2-\sin ^22x\leq 2$

Vậy $y_{\min}=1; y_{\max}=2$

i.

$-1\leq \cos (x+\frac{\pi}{6})\leq 1$

$\Rightarrow 9\leq 7-2\cos (x+\frac{\pi}{6})\leq 5$

Vậy $y_{\min}=5, y_{\max}=9$

 

Answer 2

Lời giải:

a. Vì $-1\leq \sin x\leq 1$

$\Rightarrow -1\leq 3+4\sin x\leq 7$

Vậy $y_{\min}=-1; y_{\max}=7$

b.

Vì $-1\leq \cos x\leq 1$

$\Rightarrow -7\leq 5\cos x-2\leq 3$

Vậy $y_{\min}=-7, y_{\max}=3$

c.

Vì $-1\leq \cos 2x\leq 1$

$\Rightarrow -12\leq 4\cos 2x-8\leq -4$ 

Vậy $y_{\min}=-12; y_{\max}=-4$

d.

Vì $-1\leq \sin 2x\leq 1\Rightarrow 0\leq |\sin 2x|\leq 1$

$\Rightarrow -1\leq 4-5|\sin 2x|\leq 4$

Vậy $y_{\min}=-1; y_{\max}=4$