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

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

Dễ thấy: \(f\left(x\right)=\left(x+m-1\right)^2-m^2+5m-6\ge-m^2+5m-6\)

Giá trị nhỏ nhất của f(x) đạt lớn nhất tức \(-m^2+5m-6\) đạt lớn nhất

Mà \(g\left(m\right)=-m^2+5m-6=-\left(m-\dfrac{5}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

g(m) đạt lớn nhất khi m=5/2

m cần tìm là 5/2

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

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

Hàm số \(f\left(x\right)\) liên tục trên đoạn \(\left[\frac{1}{2};2\right]\)

+)\(f'\left(x\right)=\frac{x^2+2x}{\left(x+1\right)^2};f'\left(x\right)=0\Leftrightarrow x=0\notin\left[\frac{1}{2};2\right]\)hoặc \(x=-2\notin\left[\frac{1}{2};2\right]\)

+) \(f\left(\frac{1}{2}\right)=\frac{7}{6};f\left(2\right)=\frac{7}{3}\)

Vậy \(minf\left(x\right)_{x\in\left[\frac{1}{2};2\right]}=\frac{7}{6}\) khi \(x=\frac{1}{2}\)

       \(maxf\left(x\right)_{x\in\left[\frac{1}{2};2\right]}=\frac{7}{3}\) khi \(x=2\)

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

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

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

f'(x)>0 với mọi x khác -8, suy ra hàm số đã cho đồng biến trên [0;3].

Giá trị nhỏ nhất của f(x) trên [0;3] là (-m^2)/8. Ta có: (-m^2)/8=2.

Suy ra, không có giá trị nào của số thực m thỏa yêu cầu đề bài.

sai

\(f\left(x\right)=\frac{x^2}{2}-4\ln\left(3-x\right)\) trên đoạn \(\left[-2;1\right]\)

Ta có :

         \(f'\left(x\right)=x+\frac{4}{3-x}=\frac{-x^2+3x+4}{3-x}=0\Leftrightarrow-x^2+3x+4=0\)

                                                            \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\in\left[-2;1\right]\\x=4\notin\left[-2;1\right]\end{array}\right.\)

Mà : 

   \(\begin{cases}f\left(-2\right)=2-4\ln5\\f\left(-1\right)=\frac{1}{2}-8\ln2=\frac{1-16\ln2}{2}\\f\left(1\right)=\frac{1}{2}-4\ln2=\frac{1-8\ln2}{2}\end{cases}\)    \(\Rightarrow\begin{cases}Max_{x\in\left[-2;1\right]}f\left(x\right)=\frac{1-8\ln2}{2};x=1\\Min_{x\in\left[-2;1\right]}f\left(x\right)=\frac{1-16\ln2}{2};x=-1\end{cases}\)