Question

Giải hệ phương trình

\(\left\{{}\begin{matrix}x+y+z=9\\x^2+y^2+z^2=27\end{matrix}\right.\)

Answer 1

x+y+z=9\(\Leftrightarrow\left(x+y+z\right)^2=9^2=81\Leftrightarrow x^2+y^2+z^2+2\left(xy+xz+yz\right)=81\Leftrightarrow xy+xz+yz=27\)

Mà \(x^2+y^2+z^2=27\)

Suy ra \(2\left(x^2+y^2+z^2\right)=2\left(xy+xz+yz\right)=27\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)\(\Leftrightarrow x=y=z\)

Vậy \(x=y=z=\frac{9}{3}=3\)

(x;y;z)=(3;3;3)