Question

[Võ Anh Tuấn]

Tìm các số nguyên dương x y z biết x²+y²+z²+1/x²+1/y²+1/z²=6

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Answer 1

[Pham Van Hung]

\(\Rightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)+\left(z^2+\frac{1}{z^2}-2\right)=0\)

\(\Rightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2+\left(z-\frac{1}{z}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\\z-\frac{1}{z}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\\z=\frac{1}{z}\end{cases}\Rightarrow}\hept{\begin{cases}x^2=1\\y^2=1\\z^2=1\end{cases}\Rightarrow x=y=z=1}\) (vì x,y,z > 0)