Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

a) \(y=1-2sinx\)

Ta có: \(-1\le sinx\le1\Rightarrow-2\le2sinx\le2\)

                                   \(\Rightarrow2\ge-2sin2x\ge-2\)

                                   \(\Rightarrow3\ge1-2sinx\ge-1\)

      Vậy \(y_{max}=3,y_{min}=-1\)

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 $y_{\min}=1; y_{\max}=3$

Chọn B

Đặt \(sinx+cosx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

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

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

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

\(-\dfrac{b}{2a}=-1\) 

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

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

Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

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

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

Vẽ BBT của \(f\left(t\right)=\dfrac{t^2}{2}+t-\dfrac{1}{2};t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow\)\(f\left(t\right)_{min}=-1\Leftrightarrow t=-1\Rightarrow sinx+cosx=-1\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)....

\(f\left(t\right)_{max}=\dfrac{1+2\sqrt{2}}{2}\)\(\Leftrightarrow t=\sqrt{2}\Rightarrow sinx+cosx=\sqrt{2}\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)....

`y=sin^2 5x+cos^2 5x+3sin 2x-2`       `TXĐ: R`

`y=1+3sin 2x-2`

`y=3sin 2x-1`

Ta có: `-1 <= sin 2x <= 1`

`<=>-3 <= 3sin 2x <= 3`

`<=>-4 <= y <= 2`

    `=> y_[mi n]=4<=>sin 2x =-1<=>x=-\pi/4 + k\pi`  `(k in ZZ)`

         `y_[max] = 2<=>sin 2x=1<=>x=\pi/4+k\pi`  `(k in ZZ)`