mk sửa lại đoạn sau:
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x-1=0\\2z-x-1=0\\2y+x-z-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\2z-2=0\\2y-z=0\left(x-1=0\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\2y=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\y=\dfrac{1}{2}\end{matrix}\right.\)
Ta có:
3x2+4y2+5z2+4xy-4yz-6xz-2x-4y-2z+3=0
<=> 4y2+4yx-4yz-4y+x2-2xz+z2-2x+2z+1+4z2-4xz-4z+x2+2x+1+x2-2x+1=0
<=>4y2+4y(x-z-1)+(x-z)2-2(x-z)+1+4z2-4z(x+1)+(x+1)2+(x-1)2=0
<=>4y2+4y(x-z-1)+(x-z-1)2+(2z-x-1)2+(x-1)2=0
<=>(2y+x-z-1)2+(2z-x-1)2+(x-1)2=0
<=>\(\left\{{}\begin{matrix}x-1=0\\2z-x-1=0\\2y+x-z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2z=0\left(x-1=0\right)\\2y-z=0\left(x-1=0\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\z=0\\y=0\end{matrix}\right.\)
Vay....
