Lời giải:

a. Vì $x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên $x^2+2\geq 2$

$\Rightarrow A=\frac{32}{x^2+2}\leq \frac{32}{2}=16$

Vậy $A_{\max}=16$ khi $x^2=0\Leftrightarrow x=0$

b.

$(x+1)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow 2(x+1)^2+3\geq 3$

$\Rightarrow B=\frac{5}{2(x+1)^2+3}\leq \frac{5}{3}$

Vậy $B_{\max}=\frac{5}{3}$ khi $x+1=0\Leftrightarrow x=-1$

\(C=-\left|x+\dfrac{2}{5}\right|\le0\forall x\)

\("="\Leftrightarrow x=\dfrac{-2}{5}\\ C_{max}=0\)

\(D=\dfrac{5}{17}-\left|3x-2\right|\le\dfrac{5}{17}\)

\("="\Leftrightarrow x=\dfrac{2}{3}\\ D_{max}=\dfrac{5}{17}\)

`a)C=-|x+2/5|`
`|x+2/5|>=0`
`<=>-|x+2/5|<=0`
`<=>C<=0`
Dấu "=" `<=>x=-2/5`
`b)D=5/17-|3x-2|`
`|3x-2|>=0`
`<=>5/17-|3x-2|<=5/17`
`<=>D<=5/17`
Dấu "=" `<=>x=2/3`.

a, \(cos3x^2\in\left[-1;1\right]\)

\(\Rightarrow1-cos3x^2\in\left[0;2\right]\)

\(\Rightarrow\sqrt{1-cos3x^2}\in\left[0;\sqrt{2}\right]\)

\(\Rightarrow y=\sqrt{1-cos3x^2}-2\in\left[-2;\sqrt{2}-2\right]\)

\(\Rightarrow y_{min}=-2\Leftrightarrow cos3x^2=1\Leftrightarrow3x^2=k2\pi\Leftrightarrow x=\pm\sqrt{\dfrac{k2\pi}{3}}\)

b, ĐK: \(x\ge1\)

\(cos\sqrt{x-1}\in\left[-1;1\right]\)

\(\Rightarrow y=2008cos\sqrt{x-1}\in\left[-2008;2008\right]\)

\(\Rightarrow y_{min}=-2008\Leftrightarrow cos\sqrt{x-1}=-1\Leftrightarrow\sqrt{x-1}=\pi+k2\pi\Leftrightarrow x=1+\left(\pi+k2\pi\right)^2\)

\(y_{max}=2008\Leftrightarrow cos\sqrt{x-1}=1\Leftrightarrow\sqrt{x-1}=k2\pi\Leftrightarrow x=1+4k^2\pi^2\)

Ta có: `A` lớn nhất `<=> (2015)/(18+12|x-6|)` nhỏ nhất.

`<=> 18+12|x-6|` nhỏ nhất.

`<=> 12|x-6|` nhỏ nhất, do `18` là hằng.

`<=> 12|x-6|=0`

`<=> x=6 => A=2015/18`

Vậy...

`b, B>=x+1/3+1-x`

`=4/3`.

Đẳng thức xảy ra `<=> x+1/3=1-x`

`<=> x=2/3`.

Vậy...

\(ab+bc+ca=abc\rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Có \(4A=\Sigma\frac{4}{a+b}\le\Sigma\left(\frac{1}{a}+\frac{1}{b}\right)=2\Sigma\frac{1}{a}=2\)

\(\Rightarrow A\le2\)

"=" tại a=b=c=3

a) y=\(sin^4x+cos^4x-3=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-3=-2-\dfrac{1}{2}.sin^22x\)

Có \(0\le sin^22x\le1\)

\(\Leftrightarrow-2\ge y\ge-\dfrac{5}{2}\)

Min xảy ra \(\Leftrightarrow sin^22x=1\Leftrightarrow sin2x=1\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\left(k\in Z\right)\)

\(\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\left(k\in Z\right)\)

Max xảy ra \(\Leftrightarrow sin2x=0\Leftrightarrow2x=k\Pi\Leftrightarrow x=\dfrac{k\Pi}{2}\)

b, \(x\in\left[0;\pi\right]\)

=>\(sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\dfrac{\sqrt{2}}{2};1\right]\)

\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\sqrt{2};2\right]\)

\(\Rightarrow\left\{{}\begin{matrix}Miny=-\sqrt{2}\\Maxy=2\end{matrix}\right.\)

Min xảy ra \(\Leftrightarrow x=0\)

Max xảy ra \(\Leftrightarrow x=\dfrac{\pi}{2}\)