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

\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)

\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)

\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(A\ge\dfrac{\left(x+y\right)^2}{2xy}+\dfrac{\sqrt{xy}}{x+y}\)

\(A\ge\dfrac{7\left(x+y\right)^2}{16xy}+\dfrac{\left(x+y\right)^2}{16xy}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}\)

\(A\ge\dfrac{7.4xy}{16xy}+3\sqrt[3]{\dfrac{\left(x+y\right)^2xy}{16.4.xy\left(x+y\right)^2}}=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)

Thay y= 1-x ta được

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

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

Đẳng thức xảy ra  \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y=1-x\end{cases}}\)  \(\Leftrightarrow x=y=\frac{1}{2}\)

Đặt \(x=1-y\)

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

\(\Leftrightarrow C=1-2y+y^2+y^2+y-y^2=y^2-y+1\)

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

Vậy min C là 3/4 khi y=1/2 và x =1- 1/2= 1/2 hay x=y= 1/2

Viết được bao nhiêu chữ số có 3 chữ số mà mỗi số chỉ có duy nhất 1 chữ số 4? 

mình k'o hiểu lắm . Nếu mình thì mình đã giúp bạn rồi .Cho mình xin lỗi

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

Áp dụng Cô si ,ta có:

\(x^2+\frac{1}{xy}\ge2\sqrt{\frac{x}{y}}\)

\(y^2+\frac{1}{xy}\ge2\sqrt{\frac{y}{x}}\)

\(\Rightarrow C\ge2\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\right)\ge2\)(áp dụng cô si lần nữa)

Dấu = xảy ra khi x=y=1

chắc vậy.sai thì thông cảm chị nha

Thanks

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

\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)

\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)

\(G=x^2+y^2+xy+x+y=\left[x^2+x\left(y+1\right)+\dfrac{1}{4}\left(y+1\right)^2\right]+\dfrac{3}{4}\left(y^2+\dfrac{2}{3}y+\dfrac{1}{9}\right)-\dfrac{1}{3}\)

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

\(minG=-\dfrac{1}{3}\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)