Theo đề ta có \(\left(x+\frac{1}{y}\right)\in Z\) và \(\left(y+\frac{1}{x}\right)\in Z\)\(\Rightarrow\)\(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)\in Z\)

hay \(\left(xy+\frac{1}{xy}+2\right)\in Z\)\(\Rightarrow\)\(\left(xy+\frac{1}{xy}\right)\in Z\)

Suy ra \(\left(xy+\frac{1}{xy}\right)^2\in Z\)\(\Rightarrow\)\(\left(x^2y^2+\frac{1}{x^2y^2}+2\right)\in Z\)\(\Rightarrow\)\(\left(x^2y^2+\frac{1}{x^2y^2}\right)\in Z\)

Vậy \(x^2y^2+\frac{1}{x^2y^2}\) là số nguyên (đpcm).

\(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=xy+2+\frac{1}{xy}\)

vì 2 nguyên nên \(xy+\frac{1}{xy}\)nguyên

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

nen \(x^2y^2+\frac{1}{x^2y^2}\)nguyên

Cô gái vô sinh

\(X=2\)

\(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\)

làm ơn k mk 1 cái

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{x}{y}-\frac{y}{x}\)

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

\(=\frac{x^4-x^3y+y^4-xy^3}{x^2y^2}\)

\(=\frac{x^3\left(x-y\right)-y^3\left(x-y\right)}{x^2y^2}\)

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

Em kiểm tra lại đề bài nhé  \(\frac{2}{x-y}\)hay \(\frac{2}{x-2}\)

dạ là x-y ạ

\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\le\frac{1}{16}\left(\frac{x}{x}+\frac{x}{y}+\frac{x}{x}+\frac{x}{z}\right)=\frac{1}{16}\left(2+\frac{x}{y}+\frac{x}{z}\right)\)

\(tươngtự:\frac{y}{2y+z+x}\le\frac{1}{16}\left(2+\frac{y}{z}+\frac{y}{x}\right);\frac{z}{2z+x+y}\le\frac{1}{16}\left(2+\frac{z}{x}+\frac{z}{y}\right).\text{Cộng vế theo vế ta được:}\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+y+x}\le\frac{1}{16}\left(2+2+2+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}+\frac{z}{y}\right)=\frac{1}{16}\left[6+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\right]\ge\frac{1}{16}\left(6+2\sqrt{\frac{xy}{xy}}+2\sqrt{\frac{xz}{xz}}+2\sqrt{\frac{yz}{yz}}\right)=\)

\(=\frac{12}{16}=\frac{3}{4}\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{3}{4}\left(\text{đpcm}\right)\)