Can you help me to resolve this inequality?

The denominators are x - 3 and x + 2, so to ensure a positive quantity we multiply by (x - 3)²(x + 2)² to get:

(x + 2)²(x + 2)(x - 3) < (x - 3)²(x + 2)

so, (x + 2)(x - 3)[(x + 2)² - (x - 3)] < 0

=> (x + 2)(x - 3)(x² + 3x + 7) < 0

Now, x² + 3x + 7 has no real solutions, hence x = -2 and 3 are the critical points

Testing for x < -2 we have, (x + 2)(x - 3)(x² + 3x + 7) > 0...NO

With -2 < x < 3 we have, (x + 2)(x - 3)(x² + 3x + 7) < 0...YES

For x > 3 we have, (x + 2)(x - 3)(x² + 3x + 7) > 0...NO

Therefore, -2 < x < 3 is our range.

:)>

(x+2) / (x-3) < 1 / (x+2)
[ (x+2) / (x-3) ] - [ 1 / (x+2) ] < 0
[ (x+2)^2 - (x-3) ] / [ (x-3)(x+2) ] < 0
[ x^2 + 4x + 4 - x + 3 ] / [ (x-3)(x+2) ] < 0
[ x^2 + 3x + 7 ] / [ (x-3)(x+2) ] < 0

x^2 + 3x + 7 = [ x+ (3/2) ]^2 + 7 - (9/4) = [ x+ (3/2) ]^2 + (19/4) > 0
Hence,
(x-3)(x+2) < 0
-2 < x < 3 ... Ans