Question

cho x,y,z \(\in\)R*

có: \(x^3+y^3+z^3=1\)và \(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)

tính P=\(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}\)

Answer 1

Ta có:

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

\(\Leftrightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}=-2\)

\(\Leftrightarrow x^2z+x^2y+y^2x+y^2z+z^2x+z^2y+2xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=-z\\z=-x\end{cases}}\)

Với \(x=-y\)

\(\Rightarrow x^3+y^3+z^3=1\)

\(\Rightarrow z=1\)

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

Tương tự cho các trường hợp còn lại.