Lời giải:
\(x\in [-\sqrt{2}; \sqrt{2}]\Rightarrow x^2\leq 2\Rightarrow \sqrt{x^2+1}\leq \sqrt{3}\)

\(y=\frac{x+1}{\sqrt{x^2+1}}\geq \frac{x+1}{\sqrt{3}}\geq \frac{-\sqrt{2}+1}{\sqrt{3}}\)

Vậy $y_{\min}=\frac{-\sqrt{2}+1}{\sqrt{3}}$ khi $x=-\sqrt{2}$

$y^2=\frac{x^2+2x+1}{x^2+1}=1+\frac{2x}{x^2+1}$

$y^2=2+\frac{2x-x^2-1}{x^2+1}=2-\frac{(x-1)^2}{x^2+1}\leq 2$

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

Vậy $y_{\max}=\sqrt{2}$ khi $x=1$

ĐKXĐ: \(sinx;cosx\ge0\)

Do \(\left\{{}\begin{matrix}0\le sinx\le1\\0\le cosx\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{sinx}\ge sin^2x\\\sqrt{cosx}\ge cos^2x\end{matrix}\right.\)

\(\Rightarrow\sqrt{sinx}+\sqrt{cosx}\ge sin^2x+cos^2x=1\)

\(\Rightarrow y_{min}=1\) (khi \(x=\dfrac{\pi}{2}+k2\pi\) hoặc \(k2\pi\))

Mặt khác áp dụng Bunhiacopxki:

\(y\le\sqrt{2\left(sinx+cosx\right)}\le\sqrt{2\sqrt{2\left(sin^2x+cos^2x\right)}}=\sqrt[4]{8}\)

\(y_{max}=\sqrt[4]{8}\) khi \(x=\dfrac{\pi}{4}+k2\pi\)

a)\(y=\sqrt{3}sinx+cosx=2\left(\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\right)\)\(=2\left(sinx.cos\dfrac{\pi}{6}+cosx.sin\dfrac{\pi}{6}\right)\)\(=2sin\left(x+\dfrac{\pi}{6}\right)\)

Có \(-1\le sin\left(x+\dfrac{\pi}{6}\right)\le1\) \(\Leftrightarrow-2\le2sin\left(x+\dfrac{\pi}{6}\right)\le2\)

\(\Leftrightarrow-2\le y\le2\)

miny=-2 \(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=-1\)  \(\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+2k\pi\left(k\in Z\right)\) \(\Leftrightarrow x=-\dfrac{2\pi}{3}+k2\pi\left(k\in Z\right)\)

maxy=2\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=1\) \(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\left(k\in Z\right)\)

b) \(y=sin2x-cos2x=\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\)

Có \(\sqrt{2}\ge\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\ge-\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\ge y\ge-\sqrt{2}\)

miny=\(-\sqrt{2}\) \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow2x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

maxy=\(\sqrt{2}\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{3\pi}{8}+k\pi\left(k\in Z\right)\)

c) \(y=3sinx+4cosx=5\left(\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right)\)

Đặt \(cosa=\dfrac{3}{5}\) và \(sina=\dfrac{4}{5}\)(vì cos²a+sin²a=1)

\(y=5\left(sinx.cosa+cosx.sina\right)\)\(=5sin\left(x+a\right)\)

\(\Rightarrow-5\le y\le5\)

miny=-5 <=> \(sin\left(x+a\right)=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

maxy=5 <=> \(sin\left(x+a\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

(P/s1:cái x ở câu c ấy trông nó ngu ngu??
 P/s2:sau khi load lại câu hỏi ở 1 tab khác ,thấy 1 câu trả lời nhưng vẫn đăng vì cảm thấy bỏ đi hơi phí :?)

Áp dụng quy tắc sau: Nếu \(a\sin x+b\cos y=c\Leftrightarrow a^2+b^2\ge c^2\)

a/ \(3+1\ge y^2\Leftrightarrow4\ge y^2\Leftrightarrow-2\le y\le2\)

\(y_{max}=2\Leftrightarrow\sqrt{3}\sin x+\cos x=2\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=1\Leftrightarrow\cos\dfrac{\pi}{6}.\sin x+\sin\dfrac{\pi}{6}.\cos x=1\)

\(\Rightarrow\sin\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\)

\(y_{min}=-2\Leftrightarrow\sin\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=-\dfrac{2}{3}\pi+k2\pi\)

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

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)

f(

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=t\Rightarrow t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

\(\Rightarrow y=t+\dfrac{t^2-1}{2}=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\)

Xét hàm \(y=f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(f\left(-\sqrt{2}\right)=\dfrac{1-2\sqrt{2}}{2}\) ; \(f\left(-1\right)=-1\) ; \(f\left(\sqrt{2}\right)=\dfrac{1+2\sqrt{2}}{2}\)

\(\Rightarrow y_{min}=-1\) ; \(y_{max}=\dfrac{1+2\sqrt{2}}{2}\)

2.

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

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2\left(x^2+y^2\right)+xy\ge5xy\\2\left(x^2+y^2\right)+xy\ge-3xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\ge5xy\\1\ge-3xy\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{3}\le xy\le\dfrac{1}{5}\)

Ta có:

P=\(2\left(x^2+y^2\right)^2-4x^2y^2+2+\left(x^2+y^2+2xy\right)\)

P= \(\dfrac{2\left(1-xy\right)^2}{4}-4\left(xy\right)^2+2+\left(\dfrac{1-xy}{2}+2xy\right)\)

=\(\dfrac{\left(xy\right)^2-2xy+1}{2}-4\left(xy\right)^2+2+\dfrac{3xy}{2}+\dfrac{1}{2}\)

Đặt t = xy => \(-\dfrac{1}{3}\le t\le\dfrac{1}{5}\)

Ta có : 

P= \(\dfrac{-7t^2}{2}+\dfrac{t}{2}+3=-\dfrac{7}{2}\left(t-\dfrac{1}{14}\right)^2+\dfrac{169}{56}\)

Ta có: \(-\dfrac{1}{3}-\dfrac{1}{14}\le t-\dfrac{1}{14}\le\dfrac{1}{5}-\dfrac{1}{14}\)

<=>\(-\dfrac{17}{42}\le t-\dfrac{1}{14}\le\dfrac{9}{70}\)

=> 0\(\le\left(t-\dfrac{1}{14}\right)^2\le\left(\dfrac{17}{42}\right)^2\)

\(\dfrac{169}{56}\ge P\ge\dfrac{169}{56}-\dfrac{7}{2}\left(\dfrac{17}{42}\right)^2\)

Max P= \(\dfrac{169}{56}\) => t = 1/14 => \(xy=\dfrac{1}{14}\rightarrow x^2+y^2=\dfrac{13}{14}\) => x,y=...

Min P=\(\dfrac{169}{56}-\dfrac{7}{6}\left(\dfrac{17}{42}\right)^2\) <=> \(t=xy=-\dfrac{1}{3}\)

<=> x=-y=\(\dfrac{1}{\sqrt{3}}\)