Lời giải:

a. $y=\sqrt{x^2+x-2}\geq 0$ (tính chất cbh số học)

Vậy $y_{\min}=0$. Giá trị này đạt tại $x^2+x-2=0\Leftrightarrow x=1$ hoặc $x=-2$
b.

$y^2=6+2\sqrt{(2+x)(4-x)}\geq 6$ do $2\sqrt{(2+x)(4-x)}\geq 0$ theo tính chất căn bậc hai số học

$\Rightarrow y\geq \sqrt{6}$ (do $y$ không âm)

Vậy $y_{\min}=\sqrt{6}$ khi $x=-2$ hoặc $x=4$

$y^2=(\sqrt{2+x}+\sqrt{4-x})^2\leq (2+x+4-x)(1+1)=12$ theo BĐT Bunhiacopxky

$\Rightarrow y\leq \sqrt{12}=2\sqrt{3}$

Vậy $y_{\max}=2\sqrt{3}$ khi $2+x=4-x\Leftrightarrow x=1$

c. ĐKXĐ: $-2\leq x\leq 2$

$y^2=(x+\sqrt{4-x^2})^2\leq (x^2+4-x^2)(1+1)$ theo BĐT Bunhiacopxky

$\Leftrightarrow y^2\leq 8$

$\Leftrightarrow y\leq 2\sqrt{2}$

Vậy $y_{\max}=2\sqrt{2}$ khi $x=\sqrt{2}$

Mặt khác:

$x\geq -2$

$\sqrt{4-x^2}\geq 0$

$\Rightarrow y\geq -2$
Vậy $y_{\min}=-2$ khi $x=-2$

\(y=2cos^2x-2\sqrt{3}sinx.cosx+1\)

\(=2cos^2x-1-2\sqrt{3}sinx.cosx+2\)

\(=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)

\(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)

Ta có: \(cos\left(2x+\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow min=0\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=-1\Leftrightarrow2x+\dfrac{\pi}{3}=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

\(\Rightarrow max=4\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=1\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

\(y=2cos^2x-\sqrt{3}sin2x+1=cos2x-\sqrt{3}sin2x+2\)

\(y=2.cos\left(2x+\dfrac{\pi}{3}\right)+2\)

\(\forall x\in R->-1\le cos\left(2x+\dfrac{\pi}{3}\right)\)

=> \(Min_y=2.\left(-1\right)+2=0\) 

Mặt khác, theo Bunhiacopxki:

\(\left(cos2x+\sqrt{3}sin2x\right)^2\le\left(1^2+\sqrt{3}^2\right)\left(cos^22x+sin^22x\right)=4\)

=>\(Max_y=4\)