6.

\(y=1-sin^2x+2sinx+2=-sin^2x+2sinx+3\)

\(y=-\left(sinx-1\right)^2+4\le4\)

\(y_{max}=4\) khi \(sinx=1\)

\(y=\left(sinx+1\right)\left(3-sinx\right)\ge0\)

\(y_{min}=0\) khi \(sinx=-1\)

7.

\(y=sin^4x-2\left(1-sin^2x\right)+1=sin^4x+2sin^2x-1\)

Do \(0\le sin^2x\le1\Rightarrow-1\le y\le2\)

\(y_{min}=-1\) khi \(sin^2x=0\)

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

8.

\(y=\frac{1}{3}+\frac{4}{3}cos^2x\)

Do \(0\le cos^2x\le1\Rightarrow\frac{1}{3}\le y\le\frac{5}{3}\)

\(y_{min}=\frac{1}{3}\) khi \(cos^2x=0\)

\(y_{max}=\frac{5}{3}\) khi \(cos^2x=1\)

9.

\(-1\le sin2x\le1\Rightarrow0\le1+sin2x\le2\)

\(\Rightarrow0\le y\le\sqrt{2}\)

\(y_{min}=0\) khi \(sin2x=-1\)

\(y_{max}=\sqrt{2}\) khi \(sin2x=1\)

10.

\(y=3-\left(2sinx.cosx\right)^2=3-sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow2\le y\le3\)

\(y_{min}=2\) khi \(sin^22x=1\)

\(y_{max}=3\) khi \(sin2x=0\)

12.

\(y=8+\frac{1}{4}\left(2sinx.cosx\right)=8+\frac{1}{4}sin2x\)

Do \(-1\le sin2x\le1\Rightarrow\frac{31}{4}\le y\le\frac{33}{4}\)

\(y_{min}=\frac{31}{4}\) khi \(sin2x=-1\)

\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)

13.

Về bản chất giống hệt câu 13, chỉ cần thay chữ sin bằng chữ cos

Đật 3 cái mẫu bên VT lần lượt là x,y,z rồi áp dụng C-S dạng engel

Để dễ nhìn ta đặt \(\hept{\begin{cases}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{cases}\left(a,b,c\ge0\right)}\)

Vậy BĐT đầu tương đương \(T=\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c\)

Áp dụng BĐT C-S dạng Engel ta có:

\(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{4^2}{c}\ge\frac{\left(1+2+4\right)^2}{a+b+c}=\frac{49}{a+b+c}\)

Tiếp tục dùng AM-GM ta có: \(VT\ge\frac{49}{a+b+c}+\left(a+b+c\right)\ge2\sqrt{\frac{49}{a+b+c}\cdot\left(a+b+c\right)}=2\sqrt{49}=14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=1\\b=2\\c=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}\)

nhìn qua thì chắc AM-GM+Cauchy-schwarz chắc thế :)

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 = \(\sqrt{5}\)

Đặt \(sin^24x=t\left(t\in\left[0;1\right]\right)\)

\(y=1-8sin^22x.cos^22x+2sin^42x\)

\(=1-2sin^24x+2sin^42x\)

\(\Rightarrow y=f\left(t\right)=1-2t+2t^2\)

\(y_{min}=min\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=\dfrac{1}{2}\)

\(y_{max}=max\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=1\)