Question

gấpppppppppppppppppppppppp ạ:

Answer 1

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

Ta có: \(P=\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}\)

\(=\frac{xyz}{z^3}+\frac{xyz}{x^3}+\frac{xyz}{y^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)

\(=xyz\left\lbrack\left(\frac{1}{x}+\frac{1}{y}\right)^3-3\cdot\frac{1}{x}\cdot\frac{1}{y}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}\right\rbrack\)

\(=xyz\left\lbrack\left(-\frac{1}{z}\right)^3+\frac{1}{z^3}-\frac{3}{xy}\cdot\frac{-1}{z}\right\rbrack=xyz\left\lbrack\frac{1}{-z^3}+\frac{1}{z^3}+\frac{3}{xyz}\right\rbrack=xyz\cdot\frac{3}{xyz}=3\)