Ta có:
`|2x - 3| < 5`
`<=>` \(\left[ \begin{array}{l}2x - 3 < 5\\2x - 3 > -5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x < 4\\x > -1\end{array} \right.\)
`=> -1 < x < 4`
`=>A = (-1; 4)`
`3 - |y| < 0`
`<=> |y| > 3`
`<=>` \(\left[ \begin{array}{l}y > 3\\y < -3\end{array} \right.\)
`=> B = (-∞; -3) ∪ (3; +∞)`
Vậy
`A ∩ B = (3; 4)`
`A ∪ B = (-∞; -3) ∪ (-1; +∞)`