`(x-1/2)^2=4`
`⇒(x-1/2)^2=(±2)^2`
`⇒`\(\left[ \begin{array}{l}x-\frac{1}{2}=2\\x-\frac{1}{2}=-2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{-3}{2}\end{array} \right.\)
Vậy `x∈{5/2;-3/2}`
`x^2 =16`
`⇒x^2=(±4)^2`
`⇒` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy `x∈{±4}`