\(f\left(x\right)=3x+\frac{2}{\left(2x+1\right)^2}=\frac{3}{4}\left(2x+1\right)+\frac{3}{4}\left(2x+1\right)+\frac{2}{\left(2x+1\right)^2}-\frac{3}{2}\)

\(\ge3\sqrt[3]{\left[\frac{3}{4}\left(2x+1\right)\right]^2.\frac{2}{\left(2x+1\right)^2}}-\frac{3}{2}=\frac{3}{2}\sqrt[3]{9}-\frac{3}{2}\)

Dấu \(=\)khi \(\frac{3}{4}\left(2x+1\right)=\frac{2}{\left(2x+1\right)^2}\Leftrightarrow\left(2x+1\right)^3=\frac{8}{3}\Leftrightarrow x=\frac{1}{\sqrt[3]{3}}-\frac{1}{2}\).

Ta có \(f\left(x\right)-6=\dfrac{2x^3+4-6x}{x}=\dfrac{2\left(x-1\right)^2\left(x+2\right)}{x}\ge0\) nên \(f\left(x\right)\ge6\).

Đẳng thức xảy ra khi và chỉ khi x = 1.

Cách khác thì dùng AM - GM:

\(f\left(x\right)=2x^2+\dfrac{4}{x}=2x^2+\dfrac{2}{x}+\dfrac{2}{x}\ge3\sqrt[3]{2x^2.\dfrac{2}{x}.\dfrac{2}{x}}=6\).

Xảy ra đẳng thức khi x = 1.

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

Ta có: \(0\le\sin^22x\le1\)

=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)

=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)

=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)

=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)

Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

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

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)

=>sin 2x=0

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

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

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)

\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)

Ta có: \(0\le\sin^2x\le1\)

=>\(0\le6\sin^2x\le6\)

=>\(0-1\le6\sin^2x-1\le6-1\)

=>-1<=f(x)<=5

f(x) min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

f(x) max=5 khi \(\sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

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

\(f\left(x\right)=\dfrac{x^2+10x+16}{x}=x+\dfrac{16}{x}+10\ge2\sqrt{\dfrac{16x}{x}}+10=14\)

\(f\left(x\right)_{min}=14\) khi \(x=4\)

Ta có:

Khi \(x\in\left[-3;0\right]\) thì \(f\left(x\right)\in\left[-4;5\right]\) (dùng BBT)

Lại có:

\(y=f\left(f\left(x\right)\right)=f^2\left(x\right)+6f\left(x\right)+5\) 

Khi \(f\left(x\right)\in\left[-4;5\right]\) thì \(f\left(f\left(x\right)\right)\in\left[-4;60\right]\) (dùng BBT)

Do đó, \(m=-4\Leftrightarrow f\left(x\right)=-3\Leftrightarrow x=-2\)

và \(M=60\Leftrightarrow f\left(x\right)=5\Leftrightarrow x=0\)

\(\Rightarrow S=m+M=-4+60=56\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left\lbrack\frac12\cdot2\cdot\sin x\cdot cosx\right\rbrack^2=5-2\cdot\left\lbrack\frac12\cdot\sin2x\right\rbrack^2\)

\(=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

\(0\le\sin^22x\le1\)

=>\(0\ge-\frac12\sin^22x\ge-\frac12\)

=>\(0+5\ge-\frac12\sin^22x+5\ge-\frac12+5\)

=>\(5\ge-\frac12\sin^22x+5\ge\frac92\)

=>\(\frac92\le-\frac12\sin^22x+5\le5\)

=>\(\sqrt{\frac92}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{2}{3\sqrt2}\ge\frac{1}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1}{\sqrt5}\)

=>\(\frac{2\cdot4}{3\sqrt2}\ge\frac{1\cdot4}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1\cdot4}{\sqrt5}\)

=>\(\frac{4\sqrt2}{3}\ge y\ge\frac{4}{\sqrt5}\)

=>\(y_{\max}=\frac{4\sqrt2}{3}\) khi \(-\frac12\cdot\sin^22x+5=\frac92\)

=>\(-\frac12\cdot\sin^22x=-\frac12\)

=>\(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

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

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4}{\sqrt5}\) khi \(-\frac12\cdot\sin^22x+5=5\)

=>\(\sin^22x=0\)

=>sin 2x=0

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

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

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\left(1-cos^2x\right)+5\cdot cos^2x-4\left(2\cdot cos^2x-1\right)-2\)

\(=3-3\cdot cos^2x+5\cdot cos^2x-8\cdot cos^2x+4-2=-6\cdot cos^2x+5\)

Ta có: \(0<=cos^2x\le1\)

=>\(0\ge-6\cdot cos^2x\ge-6\)

=>\(0+5\ge-6\cdot cos^2x+5\ge-6+5\)

=>5>=y>=-1

Do đó: \(y_{\min}=-1\) khi \(-6\cdot cos^2x+5=-1\)

=>\(-6\cdot cos^2x=-6\)

=>\(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

y max=5 khi \(-6\cdot cos^2x+5=5\)

=>\(-6\cdot cos^2x=0\)

=>cosx=0

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