#)Giải :

Ta có : \(x^2+y^2-xy=4\Leftrightarrow x^2+y^2=4+xy\Leftrightarrow3\left(x^2+y^2\right)=8\left(x+y\right)^2\ge8\)

\(\Rightarrow A_{max}=8\)

Dấu''='' xảy ra khi x = y = 2 hoặc x = y = -2

\(=>x^2+y^2-xy=4=x^2+y^2=4+xy=3\left(x^2+y^2\right)=8\left(x+y\right)^2>8\)

\(=>A=8\)

~Study well~ :)

\(A=\dfrac{x^3+y^3+4}{xy+1}\ge\dfrac{x^3+y^3+4}{\dfrac{x^2+y^2}{2}+1}=\dfrac{x^3+y^3+4}{2}=\dfrac{\dfrac{1}{2}\left(x^3+x^3+1\right)+\dfrac{1}{2}\left(y^3+y^3+1\right)+3}{2}\)

\(\ge\dfrac{\dfrac{3}{2}\left(x^2+y^2\right)+3}{2}=3\)

\(A_{min}=3\) khi \(x=y=1\)

Do \(x^2+y^2=2\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{2}\\y\le\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3\le\sqrt{2}x^2\\y^3\le\sqrt{2}y^2\end{matrix}\right.\)

\(\Rightarrow A\le\dfrac{\sqrt{2}\left(x^2+y^2\right)+4}{xy+1}=\dfrac{4+2\sqrt{2}}{xy+1}\le\dfrac{4+2\sqrt{2}}{1}=4+2\sqrt{2}\)

\(A_{max}=4+2\sqrt{2}\) khi \(\left(x;y\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)

tích mình với

ai tích mình

mình tích lại

thanks

Tích mình đi mình tích lại

A=x³+y³=(x+y)(x²-xy+y²)

=(x+y)²\(\ge\)0

Dấu "=" xảy ra khi x=-y

*Max

2(x^2+y^2)-2xy=8

<=> x^2+y^2+ (x-y)^2=8

<=> A\(\le\)8

Dấu bằng xảy ra khi (x,y)={(2,2),(-2,-2)}

*Min

2(x^2+y^2)=8+2xy

<=>3(x^2+y^2)=8+x^2+y^2+2xy

<=>3A=8+(x+y)^2

<=>A\(\ge\)8/3

Dấu bằng xảy ra khi (x,y)={(\(\frac{\sqrt{2}}{3},-\frac{\sqrt{2}}{3}\)),(\(-\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3}\))}

Thanks bn nha

ko có j