`d, Q(x)=(x-2)^2+(x^2-4)^2`
`⇔(x-2)^2+(x^2-4)^2=0`
`⇔(x-2)^2+[(x-2)(x+2)]^2=0`
`⇔(x-2)^2+(x-2)^2.(x+2)^2=0`
`⇔(x-2)^2[1+(x+2)^2]=0`
`⇔(x-2)^2[1+(x+2)^2]=0`
`⇔TH1:(x-2)^2=0⇔x-2=0⇔x=2`
`⇔TH2:1+(x+2)^2=0⇔(x+2)^2=-1`
$\text{Mà $(x+2)^2$ ≥ 0 ∀ x}$
`⇒ x∈∅`
$\text{Vậy x = 2}$