x+y+z=-2 Mk làm rùi 

x² + y² + z² + 2x - 4y + 6z = -14

=> x² + y² + z² + 2x - 4y + 6z +14=0

=>(x²+2x+1)+(y²-4y+4)+(z²+6z+9)=0

=>(x+1)²+(y-2)²+(z+3)²=0

Ta thấy: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y-2\right)^2\ge0\\\left(z+3\right)^2\ge0\end{cases}}\)

=>(x+1)²+(y-2)²+(z+3)^2\(\ge\)0

\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-1\\y=2\\z=-3\end{cases}}\)

\(\Rightarrow x+y+z=\left(-1\right)+2+\left(-3\right)=-2\)

chuyển vế tách HĐT tính được x=-1,y=2;z=-3 nên x+y+z=-2

\(x^2+y^2+z^2+2x-4y+6z=-14\\ x^2+2x+1+y^2-4y+4+z^2+6z+9=0\\ \left(x+1\right)^2+\left(y-4\right)^2+\left(z+3\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}x+1=0\\y-4=0\\z+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=4\\z=-3\end{matrix}\right.\\ \Rightarrow x+y+z=-1+4-3=0\)

có \(x^2+y^2+z^2+2x-4y+6z=-14\)

=>\(x^2+z^2+y^2+2x-4y+6z+14=0\)

=>\(\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6y+9\right)=0\)

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

=> x+1 =0

y-2 =0

z+3 =0

=> x = -1

y = 2

z = -3

=> x + y + z = -2

Đề bài yêu cầu gì vậy em.

\(x^2+y^2+z^2+2x-4y+6z=-14\)

\(x^2+y^2+z^2+2x-4y+6z+14=0\)

\(x^2+2x+1+y^2-4y+4+z^2+6z+9=0\)

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

vậy x + y + z = -1 + 2 + (-3) = -2

a) \(x^2+xy+y^2+1\)

\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)

\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)

mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)

\(\Rightarrow dpcm\)

b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)

\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)

\(\Rightarrow dpcm\)