Solve x2−2x−8x2−1≤0\frac{x^2 - 2x - 8}{x^2 - 1} \leq 0x2−1x2−2x−8≤0.
[−2,−1)∪(1,4][ -2, -1 ) \cup ( 1, 4 ][−2,−1)∪(1,4]
(−1,1)∪[−2,4]( -1, 1 ) \cup [ -2, 4 ](−1,1)∪[−2,4]
(−2,4)( -2, 4 )(−2,4)
(−∞,−2]∪[4,∞)( -\infty, -2 ] \cup [ 4, \infty )(−∞,−2]∪[4,∞)