`a) ( x ^2 - 1 ) . ( 2y + 4 ) = 0`
`⇒` \(\left[ \begin{array}{l}x^2-1=0\\2y + 4=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x^2=1\\2y =-4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1\\y=-2\end{array} \right.\)
Vậy `x = 1` ; `y=-2`
`b)` `| 2x - 1| - 3 = 0`
`=> | 2x - 1 | = 3`
`=>` \(\left[ \begin{array}{l}2x - 1 =3\\2x - 1 = -3\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x = 4\\2x = -2\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x = 2\\x=-1\end{array} \right.\)
Vậy `x ∈ { 2 ; -1 }`
