\(x^2+y^2+\text{z}^2+1+1+1-2x-2y-2\text{z}=0\\ \left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(\text{z}^2-2\text{z}+1\right)=0\\ \left(x-1\right)^2+\left(y-1\right)^2+\left(\text{z}-1\right)^2=0=>x-1=0\\ y-1=0\\ \text{z}-1=0\\ =>x=y=\text{z}=1\)
`x^2 + y^2 + z^2 - 2x - 2y - 2x + 3 = 0`
`<=> x^2 - 2x +1 + y^2 - 2y + 1 + z^2 - 2z + 1 = 0`
`<=> (x-1)^2 + (y-1)^2 + (z-1)^2 = 0`
Mà `(x-1)^2, (y-1)^2, (z-1)^2 >=0 forall x in RR`.
`=>` Dấu bằng xảy ra `<=> {(x-1 = 0),(y-1=0), (z-1=0):}`
`<=> {(x=1), (y=1), (z=1):}`.
Vậy `x, y, z = 1`.
\(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow x=y=z=1\)
