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

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

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

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

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

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

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

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

x+y+z=-2 Mk làm rùi 

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

Ta có:

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

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

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

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+\left(z^2-6z+9\right)=0\)\(\Leftrightarrow\left(x^2-2x.1+1^2\right)+\left(y^2+2y.2+2^2\right)+\left(z^2-2z.3+3^2\right)=0\)

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

Lại có:

\(\left(x+1\right)^2\ge0\)

\(\left(y+2\right)^2\ge0\)

\(\left(z-3\right)^2\ge0\)

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

Dấu "=" chỉ xảy ra khi và chỉ khi \(x-1=y+2=z-3=0\)

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

Khi đó: \(x+y+z=1-2+3=2\)

Ta có: \(\frac{x}{-3}=\frac{y}{7}\Leftrightarrow\frac{x}{6}=\frac{y}{-14}\) mà \(\frac{y}{-14}=\frac{z}{5}\)

=> \(\frac{x}{6}=\frac{y}{-14}=\frac{z}{5}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có: 

\(\frac{x}{6}=\frac{y}{-14}=\frac{z}{5}=\frac{2x+4y-6z}{12-56-30}=-\frac{15}{74}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{15}{74}\cdot6=-\frac{45}{37}\\y=-\frac{15}{74}\cdot\left(-14\right)=\frac{105}{37}\\z=-\frac{15}{74}\cdot-\frac{75}{74}\end{cases}}\)

x² + y² +z² + 2x - 4y+6z + 14=0

(x² + 2x +1) + (y² - 2.y.2 +2²) + (z² + 2.z.3 +3²) =0

(x+1)² + (y-2)² +(z+3)² =0

vì (x+1)² >= 0; (y-2)²>=0 ; (z+3)²>=0

nên x+1=0 và y-2=0 và z+3=0

x=-1 ; y=2 ; z=-3

vậy x+y+z=-2

2 nha bn

chuc bn hoc tot

happy new year