Answer
$\frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$
Work Step by Step
1. Factor the rational function of the numerator $\frac{9x^2+3x-2}{12x^2+5x-2}=\frac{(3x+2)(3x-1)}{(3x+2)(4x-1)}$
2. Factor the rational function of the denominator $\frac{9x^2-6x+1}{8x^2-10x-3}=\frac{(3x-1)^2}{(4x+1)(2x-3)}$
3. Combine the above results $\frac{\frac{9x^2+3x-2}{12x^2+5x-2}}{\frac{9x^2-6x+1}{8x^2-10x-3}}=\frac{(3x+2)(3x-1)}{(3x+2)(4x-1)}\times\frac{(4x+1)(2x-3)}{(3x-1)^2}=\frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$
