2х²+3х-2/(2-х)²(9-х²)>0

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To solve this inequality, we first need to find the critical points by setting the numerator and denominator equal to 0 and solving for x.

2x^2 + 3x - 2 =
(2x - 1)(x + 2) =
x = 1/2 or x = -2

(2-x)^2(9-x^2) =
(2-x)^2 = 0 or (9-x^2) =
x = 2 or x = -2 or x = 3 or x = -3

Now, we need to create intervals using the critical points and test each interval to determine the sign of the expression.

Interval 1: (-∞, -3
Choose x = -4
2(-4)^2 + 3(-4) - 2 / [(2 -(-4))^2(9-(-4)^2)
= 50 / 144 > 0

Interval 2: (-3, -2
Choose x = -2.5
2(-2.5)^2 + 3(-2.5) - 2 / [(2-(-2.5))^2(9-(-2.5)^2)
= -0.8 / 36 > 0

Interval 3: (-2, 1/2
Choose x = 0
2(0)^2 + 3(0) - 2 / [(2-0)^2(9-0^2)
= -2 / 36 < 0

Interval 4: (1/2, 2
Choose x = 1
2(1)^2 + 3(1) - 2 / [(2-1)^2(9-1^2)
= 3 / 7 > 0

Interval 5: (2, 3
Choose x = 2.5
2(2.5)^2 + 3(2.5) - 2 / [(2-2.5)^2(9-2.5^2)
= 11.2 / 9.76 > 0

Interval 6: (3, ∞
Choose x = 4
2(4)^2 + 3(4) - 2 / [(2-4)^2(9-4^2)
= 38 / 64 > 0

Therefore, the solution to the inequality 2x^2+3x-2/(2-x)^2(9-x^2) > 0 is x ∈ (-∞, -3) U (-2, 1/2) U (2, 3) U (3, ∞).