Đáp án + giải thích các bước giải:
`x^2+y^2+z^2+2(xy+yz+zx)=29+2(-10)`
`->(x+y+z)^2=9`
`->(-1+z)^2=9`
`->`\(\left[ \begin{array}{l}z-1=3\\z-1=-3\end{array} \right.\)
`->`\(\left[ \begin{array}{l}z=4\\z=-2\end{array} \right.\)
Với `z=4`
`->xy+4y+4x=-10`
`->xy+4(x+y)=-10`
`->xy+4(-1)=-10`
`->xy=-6`
mà `x+y=-1->x=-1-y`
`->(-1-y)y=-6`
`->-y^2-y=-6`
`->y^2+y-6=0`
`->y^2-2y+3y-6=0`
`->y(y-2)+3(y-2)=0`
`->(y+3)(y-2)=0`
`->`\(\left[ \begin{array}{l}y+3=0\\y-2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}y=-3\\y=2\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-1-(-3)=2\\x=-1-2=-3\end{array} \right.\)
`-> (x;y;z)=(2;-3;4);(-3;2;4)`
Với `z=-2 `
`->xy-2y-2x=-10`
`->xy-2(x+y)=-10`
`->xy-2(-1)=-10`
`->xy=-12`
mà `x+y=-1->x=-1-y`
`->(-1-y)y=-12`
`->-y^2-y=-12`
`->y^2+y-12=0`
`->y^2-3y+4y-12=0`
`->y(y-3)+4(y-3)=0`
`->(y+4)(y-3)=0`
`->`\(\left[ \begin{array}{l}y=-4\\y=3\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-1-(-4)=3\\x=-1-3=-4\end{array} \right.\)
`-> (x;y;z)=(3;-4;-2);(-4;3;-2)`
Vậy `(x;y;z)=(2;-3;4);(-3;2;4);(3;-4;-2);(-4;3;-2)`
