cho$:\begin{cases}x^3+y^3+z^3=1 \\ x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y})= -2\end{cases}$
Tính$:\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

ta có:
$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=-2$
$\Leftrightarrow x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=-2xyz$
$\Leftrightarrow (x+y)(y+z)(z+x)=0$
$\Rightarrow x+y=0$ hoặc:$ y+z=0$ hoặc$: z+x=0$
$\Rightarrow x=-y$ hoặc$ y=-z$ hoặc$ z=-x$
mà $:x^3+y^3+z^3=1\Rightarrow x^3+(-x)^3+z^3=1\Leftrightarrow z=1$(x;y;z hoán vị)
Nên$:\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x}-\frac{1}{x}+\frac{1}{z}=\frac{1}{z}=\frac{1}{1}=1$(x;y;z; hoán vị)
$\Rightarrow $ (x+y)(y+z)(z+x)=0
$\Rightarrow $ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$= 1

ta có:
xy+xz+yx+yz+zx+zy=2
x2y+x2z+y2x+y2z+z2x+z2y=2xyz
(x+y)(y+z)(z+x)=0
x+y=0 hoặc:y+z=0 hoặc:z+x=0
x=y hoặcy=z hoặcz=x
mà :x3+y3+z3=1x3+(x)3+z3=1z=1(x;y;z hoán vị)
Nên:1x+1y+1z=1x1x+1z=1z=11=1(x;y;z; hoán vị)

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