Câu hỏi của Núi non tình yêu thuần khiết

Question

Tìm x; y; z biết:

$2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0$

Dung Nguyễn Thị Xuân · Accepted Answer

$2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0$

$\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0$

$\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0$

$\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\left(x+5\right)^2=0\\left(y+3\right)^2=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}z=8\x=-5\y=-3\end{matrix}\right.$

Vậy x = -5; y = -3; z = 8Câu trả lời: $2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0$

$\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0$

$\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0$

$\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\left(x+5\right)^2=0\\left(y+3\right)^2=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}z=8\x=-5\y=-3\end{matrix}\right.$

Vậy x = -5; y = -3; z = 8