![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+z^2+2x-4y+6z=-14\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6z+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
\(\Leftrightarrow\begin{cases}x+1=0\\y-2=0\\z+3=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-1\\y=2\\z=-3\end{cases}\)
\(\Rightarrow x+y+z=-1+2-3=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(^{x^2+y^2+z^2+2x-4y+6z=-14}\)
\(=x^2+2x+1+y^2-4y+4+z^2+6z+9=-14+14=0\)\(=\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)\(\Rightarrow\left(x+1\right)^2=0;\left(y-2\right)^2=0;\left(z+3\right)^2=0\)\(\Rightarrow x+1=0;y-2=0;z+3=0\)\(\Rightarrow x=-1;y=2;z=-3\Rightarrow x+y+z=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+z^2+2x-4y-6z+14\)
\(=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2-6z+9\right)\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\)
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 \(x^2+y^2+z^2+2x-4y-6z+14\ge0\)\(\forall x,y,z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 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
=> x = -1; y = 2; z = -3
vậy x+y+z =-2