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

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

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

\(\Leftrightarrow\begin{cases}x+1=0\\y-2=0\\z+3=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-1\\y=2\\z=-3\end{cases}\)

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

x²+2x+1+y²-4y+4+z²+6z+9=0

(x+1)²+(y-2)²+(z+3)²=0

(x+1)² \(\ge0,\left(y-2\right)^2\ge0,\left(z+3\right)^2\ge0\)

mà tổng của chúng là 0 nên suy ra mỗi cái =0 nha

từ đó tính đc x,y,z

trả lời đầu tiên mk cho ko cần xét đúng sai 

Có: \(x^2+y^2+z^2+2x-4y+6z=-14\)

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

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

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

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

\(\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\)

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

Do đó: \(\hept{\begin{cases}x=-1\\y=2\\z=-3\end{cases}}\)

Vậy: \(x+y+z=-1+2-3=-2\)

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 có:

(x+1)²≥0

(y-2)²≥0

(z+3)²≥0

=>(x+1)²+(y-2)²+(z+3)²≥0

dấu "=" xảy ra khi và chỉ khi ; x+1=y-2=z+3=0

=>\(\hept{\begin{cases}x=-1\\y=2\\z=-3\end{cases}}\)

=> x+y+z=-1+2+(-3)=-2

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

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

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

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

-2

tk nhe

xin do

bye

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

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

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

Vì \(\left(x+1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\); \(\left(z-3\right)^2\ge0\forall z\)

\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

hay \(x^2+y^2+z^2+2x-4y-6z+14\ge0\)\(\forall x,y,z\)