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

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

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

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

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

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

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

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

ờ đúng ko vậy thanh nguyên chỉ có nhân =0 mới được phép tách ra chứ

Ta có: \(x^2+y^2-4x=6z-2y-z^2-14\)

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

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

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

\(\cdot\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

\(\cdot\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\)

\(\left(z-3\right)^2=0\Rightarrow z-3=0\Rightarrow z=3\)

hok tốt!

Ta có x² + y² - 4x = 6z - 2y - z² - 14

=> x² + y² - 4x - 6z + 2y + z² + 14 = 0

=> (x² - 4x + 4) + (y² + 2y + 1) + (z² - 6z + 9) = 0

=> (x - 2)² + (y + 1)² + (z - 3)² = 0

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

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y+1=0\\z-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)

Vậy x = 2 ; y = - 1 ; z = 3

x² + y² - 4x = 6z - 2y - z² - 14

<=> x² + y² - 4x - 6z + 2y + z² + 14 = 0

<=> ( x² - 4x + 4 ) + ( y² + 2y + 1 ) + ( z² - 6z + 9 ) = 0

<=> ( x - 2 )² + ( y + 1 )² + ( z - 3 )² = 0

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

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

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

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

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

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

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

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

\(\Leftrightarrow x-2=0;y+1=0;z-3=0\)

\(\Leftrightarrow x=2;y=-1;z=3\)

a: =>x^2+y^2+z^2-4x+2y-6z+14=0

=>x^2-4x+4+y^2+2y+1+z^2-6z+9=0

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

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

b: \(\left(x+y+z\right)\cdot\left(xy+yz+xz\right)\)

\(=x^2y+xyz+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2\)

\(=x^2y+xy^2+y^2z+x^2z+yz^2+xz^2+3xyz\)

Theo đề, ta có:

\(x^2y+xy^2+y^2z+x^2z+yz^2+xz^2+2xyz=0\)

\(\Leftrightarrow x^2y+2xyz+yz^2+xy^2+2xzy+xz^2+zx^2-2xyz+zy^2=0\)

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

=>x=y=z=0

=>x^2013+y^2013+z^2013=(x+y+z)^2013

#)Giải :

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

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

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

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

Vậy x = 2; y = -1; z = 3

ta có 

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

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

suy ra x=2 , y =-1 , z=3

mk nha

Cảm ơn cậu ^^ 

\(x^2+y^2+z^2=4x-2y+6z-14\Leftrightarrow x^2-4x+y^2+2y+z^2-6z+14=0\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0matkhac:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(z-3\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0mà:\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\\z=3\end{matrix}\right..Vậy:x=2;y=-1;z=3\)