\(\left\{{}\begin{matrix}x+y+z=0\left(1\right)\\y+z+t=1\left(2\right)\\z+t+x=2\left(3\right)\\t+x+y=3\left(4\right)\end{matrix}\right.\)
Cộng vế (1) (2) (3) (4) ta được :
\(x+y+z+y+z+t+z+t+x+t+x+y=0+1+2+3\)
\(\Leftrightarrow3x+3y+3z+3t=6\)
\(\Leftrightarrow3\left(x+y+z+t\right)=6\)
\(\Leftrightarrow x+y+z+t=2\)(*)
Lần lượt thay các pt (1) (2) (3) (4) vào pt (*) ta được :
\(\left\{{}\begin{matrix}0+t=2\\1+x=2\\y+2=2\\z+3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=2\\x=1\\y=0\\z=-1\end{matrix}\right.\)
Vậy (x;y;z;t) = (1;0;-1;2)
\(\left\{{}\begin{matrix}x+y+z=0\\y+z+t=1\\z+t+x=2\\t+x+y=3\end{matrix}\right.\Rightarrow x+y+z+y+z+t+z+t+x+t+x+y=6\Leftrightarrow3\left(x+y+z+t\right)=6\Leftrightarrow x+y+z+t=2.x+y+z=0\Rightarrow\left(x+y+z+t\right)-t=0\Leftrightarrow2-t=0\Leftrightarrow t=2;y+z+t=1\Leftrightarrow\left(x+y+z+t\right)-x=1\Leftrightarrow2-x=1\Leftrightarrow x=1;t+x+y=3\Leftrightarrow2+1+y=3\Leftrightarrow y=0;z+t+x=2\Leftrightarrow2+1+z=2\Leftrightarrow z=-1\)
\(Vậy:t=2;x=1;y=0;z=-1\)