Question

[Nguyễn Thành Danh]

Tìm các số x,y,z thỏa mãn:

\(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

Answer 1

[Phương Trâm]

\(x^2+y^2+z^2=xy+yz+zx\) và \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

Ta có:

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu " = " xảy ra :

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Thay \(x=y=z\) vào \(x^{2009}+y^{2009}+z^{2009}=3^{2009}\) ta được:

\(3x^{2009}=3x^{2010}\)

\(\Rightarrow x^{2009}=3^{2009}\)

\(\Rightarrow x=3\)

\(\Rightarrow y=z=x=3\)

Vậy \(\left(x;y;z\right)=\left(3;3;3\right)\)

Answer 2

[Phương Trâm]

Thiếu đề chăng.?