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13 tháng 10 2019
\(x^2+y^2+z^2=4x-2y+6z-14\)
\(\Leftrightarrow x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\y+1=0\\z-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}}\)
13 tháng 10 2019
\(\Leftrightarrow\) \(x^2\)+ \(y^2\) + \(z^2\) - \(4x\)+ \(2y\) - \(6z\) + \(14\) \(=\) \(0\)
\(\Leftrightarrow\) ( \(x^2\) - \(4x\) + \(4\) ) + ( \(y^2\) + \(2y\) + \(1\) ) \(=\) \(0\)
\(\Leftrightarrow\) ( \(x-2\))2 + \(\left(y+1\right)^2\) + \(\left(z-3\right)^2\) \(=\) \(0\)
\(\Leftrightarrow\) \(\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)
S
14 tháng 7 2019
\(x^2+y^2+z^2=4x-2y+6z-14\Leftrightarrow x^2-4x+y^2+2y+z^2-6z+14=0\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0matkhac:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(z-3\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0mà:\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\\z=3\end{matrix}\right..Vậy:x=2;y=-1;z=3\)
NT
3
9 tháng 8 2020
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!
XO
9 tháng 8 2020
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
2 tháng 9 2016
\(\Leftrightarrow x^2+y^2+z^2-4x+2y-6z+14=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)
2 tháng 9 2016
\(\Leftrightarrow x^2+y^2+z^2-4x+2y-6z+14=0\)
\(\Leftrightarrow x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
\(\Leftrightarrow x-2=0;y+1=0;z-3=0\)
\(\Leftrightarrow x=2;y=-1;z=3\)
KQ
1
28 tháng 10 2022
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
PT
1
P
26 tháng 12 2017
=(x2+2xy+y2)+(y2-4yz+4z2)+(y2-2y+1)+(z2-2z+1)-4x-2y-4z+5
=(x+y)2-4(x+y)+4 +(y-2z)2+2(y-2z)+1 +(y-1)2+(z-1)2
=(x+y-2)2+(y-2z+1)2+(y-1)2+(z-1)2\(\ge0\)\(\forall_{x,y,z}\)
Lai co (x+y-2)2+(y-2z+1)2+(y-1)2+(z-1)2\(\le\)0
=> (x+y-2)2+(y-2z+1)2+(y-1)2+(z-1)2=0
Dau = xay ra khi x=y=z=1
Đề đúng
\(x^2+y^2+z^2=4x-2y+6z-14\)
\(\Leftrightarrow x^2+y^2+z^2-4x+2y-6z+14=0\)
\(\Leftrightarrow x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
\(\Leftrightarrow x-2=0;y+1=0;z-3=0\)
\(\Leftrightarrow x=2;y=-1;z=3\)