\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?

A= \(\frac{1}{\left(x+y\right)\left(x^2+y^2-xy\right)+xy}+\frac{4x^2y^2+2}{xy}=\)\(\frac{1}{x^2+y^2}+4xy+\frac{2}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+4xy+\frac{1}{4xy}+\frac{5}{4xy}\) (1)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};a+b\ge2\sqrt{ab},\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)áp dụng vào trên ta được

 (1) \(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{5}{4}.\frac{4}{\left(x+y\right)^2}=4+2+\frac{5}{4}.4=11.\)

dấu '=" khi x=y = 1/2

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

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

\(A_{min}=3\) 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 \(\left(x;y\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)

pro rồi thì bạn cần gì mình giải nhỉ

??

\(A=x-2y+3\Rightarrow x=A+2y-3\)

\(\Rightarrow\left(2y+A-3\right)^2+y\left(A+2y-3\right)+2y^2=1\)

\(\Leftrightarrow8y^2+\left(5A-15\right)y+A^2-6A+8=0\)

\(\Delta=\left(5A-15\right)^2-32\left(A^2-6A+8\right)\ge0\)

\(\Leftrightarrow-7A^2+42A-31\ge0\)

\(\Rightarrow\dfrac{21-4\sqrt{14}}{7}\le A\le\dfrac{21+4\sqrt{14}}{7}\)

Ta có:

\(\left(x+y+1\right)xy=x^2+y^2\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{x^2}+\frac{1}{y^2}\)

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

\(\Leftrightarrow0\le\frac{1}{x}+\frac{1}{y}\le4\)

Ta lại có:

\(\frac{1}{x^3}+\frac{1}{y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)

PS: Sửa đề tìm max nhé

A=x³+y³=(x+y)(x²-xy+y²)

=(x+y)²\(\ge\)0

Dấu "=" xảy ra khi x=-y