Question

1) Tìm x, y , z . Thõa mãn phương trình sau:

9x²+y²+2z²-18x +4z-6y+20=0

Answer 1

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left[\left(3x\right)^2-18x+9\right]+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y^2-3\right)^2+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

Vậy ..........