Ta có
\(x^2+y^2+z^2\ge xy+yz+zx\)vì\(2\left(x^2+y^2+z^2\right)\ge2\left(xy-yz-zx\right)\)(biết hằng thức là tự hiểu) \(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(luôn đúng)
dấu "="xảy ra:x=y=z\(\Rightarrow x^{2009}+y^{2009}+z^{2009}=3^{2010}\)\(\Leftrightarrow3.x^{2010}=3^{2010}\)\(\Leftrightarrow x^{2010}=1\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
OK???????
Từ \(x^2+y^2+z^2=xy+yz+zx\)
=> \(2x^2+2y^2+2z^2=2xy+2yz+2xz\) (nhân cả hai vế với 2)
=> \(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
=> \(\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+\left(y^2-2yz+z^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\Rightarrow x-y=0;x-z=0;y-z=0\)
=> x = y = z thay vào \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)
=> \(3x^{2009}=3y^{2009}=3z^{2009}=3^{2010}\)
=> x = y = z = 3
Vậy x=y=z =3
kinh khủng vậy ?
x2+y2+z2 =xy+yz+zx
<=>x^2+y^2+z^2-xy-yz-zx =0
<=>2(x^2+y^2+z^2-xy-yz-zx) =0 ( nhân 2 vế cho2)
<=>2x2 + 2y2 +2z2 - 2xy - 2yz - 2zx =0
<=> x^2 -2xy+y^2 + x^2 - 2zx - z^2+y^2-2yz+z^2
=0
<=> (x-y)^2 + ( x-z)^2+(y-z)^2 =0
<=> x=y=z