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...

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 \(\Leftrightarrow x=\dfrac{\pi}{2}\)

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 $x+1=0\Leftrightarrow x=-1$

Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Đặt a+b=x;b+c=y;c+a=z

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)

Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)

2, rút gọn B=x^2/(y-1)+y^2/(x-1) 

AM-GM : x^2/(y-1)+4(y-1) >/ 4x ; y^2/(x-1)+4(x-1) >/ 4y 

=> B >/ 4x-4(y-1)+4y-4(x-1)=4x-4y+4+4y-4x+4=8 

minB=8 

Câu 1:

Áp dụng BĐT AM-GM ta có: \(x+1\ge2\sqrt{x}\)

\(\Rightarrow x+1+x+1\ge x+2\sqrt{x}+1\)

\(\Rightarrow2x+2\ge\left(\sqrt{x}+1\right)^2\left(1\right)\)

Tương tự cũng có: \(2y+2\ge\left(\sqrt{y}+1\right)^2\left(2\right)\)

Nhân theo vế của \(\left(1\right);\left(2\right)\) ta có:

\(\left(2x+2\right)\left(2y+2\right)\ge\left(\sqrt{x}+1\right)^2\left(\sqrt{y}+1\right)^2\ge16\)

\(\Rightarrow4\left(x+1\right)\left(y+1\right)\ge16\Rightarrow\left(x+1\right)\left(y+1\right)\ge4\)

Lại áp dụng BĐT AM-GM ta có:

\(\left(x+1\right)+\left(y+1\right)\ge2\sqrt{\left(x+1\right)\left(y+1\right)}\ge4\)

\(\Rightarrow x+y\ge2\). Giờ thì áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(A=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Đẳng thức xảy ra khi \(x=y=1\)

x,y có dương đâu mà AM-GM rồi schwarz hay vậy Thắng ? 

Gọi cái biểu thức đó là P nha

Trước tiên chứng minh:

\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\left(\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right)=0\)

\(\Leftrightarrow\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\Leftrightarrow x-y+y-z+z-x=0\)( đúng )

Giờ ta quay lại bài toán ban đầu 

Ta có:

\(\Leftrightarrow2P=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)

\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)

\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)

\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}=\frac{1}{2}\)

\(\Rightarrow P\ge\frac{1}{4}\)