`x^2 - 2x + y^2 + 4y + 5 + (2z - 3)^2 = 0`
`=> (x^2 - 2x + 1) + (y^2 + 4y + 4) + (2z - 3)^2 = 0`
Ta có: ` x^2 - 2x + 1 = x^2 - x - x + 1 = x. (x - 1) - (x - 1) = (x - 1)^2`
`y^2 + 4y + 4 = y^2 + 2y + 2y + 4 = y(y + 2) + 2(y + 2) = (y + 2)^2`
`=> (x - 1)^2 + (y + 2)^2 + (2z - 3)^2 = 0`
Vì \(\left\{\begin{matrix}(x - 1)^2 \geq 0\forall x\\(y+2)^2 \geq 0 \forall y\\(2z-3)^2\geq0\forall z\end{matrix}\right.\)
`=> (x - 1)^2 + (y+2)^2 + (2z-3)^2 >= 0 \forall x, y, z`
Dấu `"="` xảy ra
`<=>` \(\left\{\begin{matrix}x - 1=0\\y+2=0\\2z-3=0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x=1\\y=-2\\z=1,5\end{matrix}\right.\)
Vậy `x = 1; y = -2; z = 1,5`
