Ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{1}{z}^3\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+3\cdot\frac{1}{x}\cdot\frac{1}{y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{z^3}\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3\cdot\frac{1}{x}\cdot\frac{1}{y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{3}{xyz}.\)Vì \(\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\)
Mặt khác : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+zx}{xyz}=0\)
\(\Rightarrow xy+yz+zx=0\)`
\(A=\frac{yz}{x^2}+2yz+\frac{xz}{y^2}+2xz+\frac{xy}{z^2}+2xy\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}+2\left(xy+yz+xz\right)\)Vì x , y , z khác 0 .
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)Vì \(xy+yz+xz=0\)
\(=xyz\cdot\frac{3}{xyz}\)Vì \(\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{3}{xyz}\)
\(=3\)
Vậy \(A=3\)
mk tưởng chố \(\left(\frac{1}{x}+\frac{1}{y}\right)^3\)phải bằng\(\left(\frac{-1}{z}\right)^3\)chứ
