Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(P=\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=\frac{x^3y^3+y^3z^3+x^3z^3}{x^2y^2z^2}\)
Áp dụng nếu a+b+c=0 thì a3+b3+c3=3abc
Với a=xy, b=yz, c=zx
Ta có: \(P=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)
Vậy P=3