`(x^2 - 4)(x^2 - 9)(x + 5) = 0`
`<=>` \(\left[ \begin{array}{l}x = ±2\\x = ±3\\x = -5\end{array} \right.\)
`=> A = {-5; -3; -2; 2; 3}`
`2x^2 - 3x - 5 = 0`
`<=>` \(\left[ \begin{array}{l}x = -1\\x = \dfrac{5}{2}\end{array} \right.\)
`=> B = {-1; 5/2}`
`3 < x <= 10`
`=> C = {4; 5; 6; 7; 8; 9; 10}`
Ta có:
`A ∩ B = ∅`
`A ∩ C = ∅`
`B ∩ C = ∅`
`A ∪ B = {-5; -3; -2; -1; 2; 5/2; 3}`
`A ∪ C = {-5; -3; -2; 2; 3; 4; 5; 6; 7; 8; 9; 10}`
`B ∪ C = {-1; 5/2; 4; 5; 6; 7; 8; 9; 10}`
`A \\ B = {-5; -3; -2; 2; 3}`
`A \\ C = {-5; -3; -2; 2; 3}`
`B \\ C = {-1; 5/2}`
