Đặt \(\left\{{}\begin{matrix}\sqrt{5sin^2x+1}=a\\\sqrt{5cos^2x+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}1\le a;b\le\sqrt{6}\\a^2+b^2=5\left(sin^2x+cos^2x\right)+2=7\end{matrix}\right.\)

\(y=a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{14}\)

\(y_{max}=\sqrt{14}\) khi \(cos2x=0\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Do \(1\le a\le\sqrt{6}\Rightarrow\left(a-1\right)\left(a-\sqrt{6}\right)\le0\)

\(\Rightarrow a\ge\dfrac{a^2+\sqrt[]{6}}{\sqrt{6}+1}\)

Tương tự ta có \(b\ge\dfrac{b^2+\sqrt{6}}{\sqrt{6}+1}\)

\(\Rightarrow y=a+b\ge\dfrac{a^2+b^2+2\sqrt{6}}{\sqrt{6}+1}=\dfrac{7+2\sqrt{6}}{\sqrt{6}+1}=\sqrt{6}+1\)

\(y_{min}=\sqrt{6}+1\) khi \(sin2x=0\Rightarrow x=\dfrac{k\pi}{2}\)

\(y\le\sqrt{2\left(6-2x+3+2x\right)}=3\sqrt{2}\)

\(y_{max}=3\sqrt{2}\) khi \(x=\dfrac{3}{4}\)

\(y\ge\sqrt{6-2x+3+2x}=3\)

\(y_{min}=3\) khi \(\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Lời giải:
ĐKXĐ: \(x\leq \frac{3}{2}\)

Hàm số chỉ có min chứ không có max bạn nhé.

\(y=\sqrt{3-2x}+\sqrt{5-2x}\)

\(\Rightarrow y^2=3-2x+5-2x+2\sqrt{(3-2x)(5-2x)}\)

\(=8-4x+2\sqrt{(3-2x)(5-2x)}\)

Ta thấy:
Vì \(x\leq \frac{3}{2}\Rightarrow 8-4x\geq 8-4.\frac{3}{2}=2\)

\(2\sqrt{(3-2x)(5-2x)}\geq 0\) (theo tính chất căn bậc 2)

\(\Rightarrow y^2=8-4x+2\sqrt{(3-2x)(5-2x)}\geq 2\)

\(\Rightarrow y\geq \sqrt{2}\) (do $y$ không âm)

Vậy $y_{\min}=\sqrt{2}$ khi $x=\frac{3}{2}$

Em mới học dạng này sơ sơ thôi nên không rành lắm, mọi người check giúp ạ.

ĐK x =< 3/2

Xét \(x_1< x_2\le\frac{3}{2}\)

\(y=f\left(x\right)=\sqrt{3-2x}+\sqrt{5-2x}\)

Ta có: \(f\left(x_1\right)-f\left(x_2\right)=\left(\sqrt{3-2x_1}-\sqrt{3-2x_2}\right)+\left(\sqrt{5-2x_1}-\sqrt{5-2x_2}\right)>0\)(do dễ thấy(em lười viết ra quá) rằng mỗi cái ngoặc đều lớn hơn 0)

Do đó f(x1) > f(x2). Do vậy x càng tăng thì giá trị f(x) càng nhỏ hay y đạt cực tiểu tại x = 3/2. Vậy \(y_{min}=\sqrt{3-2.\frac{3}{2}}+\sqrt{5-2.\frac{3}{2}}=\sqrt{2}\)

Đẳng thức xảy ra khi x = 3/2

Vậy...

a, \(y=2sin^2x-cos2x=1-2cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=2sin^2x-cos2x\in\left[-1;3\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=-1\\y_{max}=3\end{matrix}\right.\)

1: \(-1<=cosx\le1\)

=>\(-3\le-3\cdot cosx\le3\)

=>\(-3+5\le-3\cdot cosx+5\le3+5\)

=>2<=y<=8

y min=2 khi cosx=1

=>\(x=k2\pi\)

y min=8 khi cosx=-1

=>\(x=\pi+k2\pi\)

3: \(y=cos^2x+2\cdot cos2x\)

\(=\frac{1+cos2x}{2}+2\cdot cos2x=2,5\cdot cos2x+0,5\)

Ta có: \(-1\le cos2x\le1\)

=>\(-2,5\le2,5cos2x\le2,5\)

=>\(-2,5+0,5\le2,5cos2x+0,5\le2,5+0,5\)

=>-2<=y<=3

y min=-2 khi cos2x=-1

=>\(2x=\pi+k2\pi\)

=>\(x=\frac{\pi}{2}+k\pi\)

y max=3 khi cos2x=1

=>\(2x=k2\pi\)

=>\(x=k\pi\)

6: \(y=\sqrt3\cdot\sin x-cosx-2\)

\(=2\left(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx\right)-2=2\cdot\sin\left(x-\frac{\pi}{6}\right)-2\)

Ta có: \(-1\le\sin\left(x-\frac{\pi}{6}\right)\le1\)

=>\(-2\le2\sin\left(x-\frac{\pi}{6}\right)\le2\)

=>\(-2-2\le2\sin\left(x-\frac{\pi}{6}\right)-2\le2-2\)

=>-4<=y<=0

y min=-4 khi \(\sin\left(x-\frac{\pi}{6}\right)=-1\)

=>\(x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{2}+\frac{\pi}{6}+k2\pi=-\frac26\pi+k2\pi=-\frac13\pi+k2\pi\)

y max=0 khi \(\sin\left(x-\frac{\pi}{6}\right)=1\)

=>\(x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac23\pi+k2\pi\)

a/ \(0\le cos^2x\le1\Rightarrow2\le y\le\sqrt{7}\)

\(y_{min}=2\) khi \(cos^2x=1\)

\(y_{max}=\sqrt{7}\) khi \(cos^2x=0\)

b/ \(y=\frac{2}{1+tan^2x}=\frac{2}{\frac{1}{cos^2x}}=2cos^2x\le2\)

\(\Rightarrow y_{max}=2\) khi \(cos^2x=1\)

\(y_{min}\) ko tồn tại

c/ \(y=1-cos2x+\sqrt{3}sin2x=2\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+1\)

\(y=2sin\left(2x-\frac{\pi}{6}\right)+1\)

Do \(-1\le sin\left(2x-\frac{\pi}{6}\right)\le1\Rightarrow-1\le y\le3\)