Ta có : \(x^3+y^3=2x^2y^2\Rightarrow\left(x^3+y^3\right)^2=4x^4y^4\)

            \(x^6+y^6+2x^3y^3=4x^4y^4\Rightarrow x^6+y^6-2x^3y^3=4x^4y^4-4x^3y^3\)

            \(\left(x^3-y^3\right)^2=4x^3y^3\left(xy-1\right)\Rightarrow xy-1=\frac{\left(x^3-y^3\right)^2}{4x^3y^3}\)

            \(\frac{xy-1}{xy}=\frac{\left(x^3-y^3\right)^2}{4x^4y^4}\) (chia cả 2 vế cho xy)\(\Rightarrow1-\frac{1}{xy}=\frac{\left(x^3-y^3\right)^2}{4x^4y^4}\)

              \(\Rightarrow\sqrt{1-\frac{1}{xy}}=\frac{x^3-y^3}{2x^2y^2}\)

nhớ k mình nha

\(x^3+y^3=2x^2y^2\)

<=>   \(\left(x^3+y^3\right)^2=4x^4y^4\)

<=>  \(\left(x^3-y^3\right)^2=4x^4y^4-4x^3y^3\)

<=>  \(\left(x^3-y^3\right)^2=4x^4y^4\left(1-\frac{1}{xy}\right)\)

<=>  \(1-\frac{1}{xy}=\frac{\left(x^3-y^3\right)^2}{4x^4y^4}\)

<=>  \(\sqrt{1-\frac{1}{xy}}=\frac{\left|x^3-y^3\right|}{2x^2y^2}\) là số hữu tỉ

1/x+1/y+1/z =0 nhé

\(\sqrt{x^2+y^2+z^2}=\sqrt{\left(x+y+z\right)^2-2\left(xy+yz+xz\right)}=\sqrt{\left(x+y+z\right)^2-2xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=\sqrt{\left(x+y+z\right)^2}=\left|x+y+z\right|\)

Vì \(x\ne0,y\ne0\) nên điều kiện đã cho tương đương với \(\frac{x}{y^2}+\frac{y}{x^2}=2\Rightarrow\frac{x^2}{y^4}+\frac{y^2}{x^4}+\frac{2}{xy}=4\Leftrightarrow4\left(1-\frac{1}{xy}\right)=\frac{x^2}{y^4}+\frac{y^2}{x^4}-\frac{2}{xy}=\left(\frac{x}{y^2}-\frac{y}{x^2}\right)^2\)

\(\Rightarrow\sqrt{1-\frac{1}{xy}}=\frac{1}{2}\left|\frac{x}{y^2}-\frac{y}{x^2}\right|\)