Question

giải phương trình x² + y² + z² = y(x + z )

Answer 1

\(x^2+y^2+z^2=y\left(x+z\right)\\ < =>x^2+y^2+z^2=xy+yz\\ < =>2x^2+2y^2+2z^2=2xy+2yz\\ < =>x^2-2xy+y^2+y^2-2yz+z^2+x^2+z^2=0\\ < =>\left(x-y\right)^2+\left(y-z\right)^2+x^2+z^2=0\left(1\right)\)

ta thấy

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\ \left(y-z\right)^2\ge0\forall y,z\\x^2\ge0\forall x\\z^2\ge0\forall z\end{matrix}\right.\)

mà để (1) luôn đúng

\(< =>\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\x^2=0\\z^2=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x-y=0\\y-z=0\\x=0\\z=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

vậy x = y = z = 0

chúc may mắn :)