Answer
$\frac{11x^2-x-11}{(2x-1)(x+3)(3x+2)}$, $x\ne-3, -\frac{2}{3},\frac{1}{2}$.
Work Step by Step
Step 1. The domain requirements of the rational expressions are that $2x^2+5x-3\ne0$, which gives $(2x-1)(x+3)\ne0$ and $x\ne-3,\frac{1}{2}$. We also need $6x^2+x-2\ne0$, which gives $(2x-1)(3x+2)\ne0$ and $x\ne-\frac{2}{3},\frac{1}{2}$. Thus, we have $x\ne-3, -\frac{2}{3},\frac{1}{2}$.
Step 2. Factor the numerator and the denominator completely. We have $\frac{4x-1}{(2x-1)(x+3)}-\frac{x+3}{(2x-1)(3x+2)}$
Step 3. We can find the least common denominator as $LCD=(2x-1)(x+3)(3x+2)$ and we can calculate the original expression as $\frac{(4x-1)(3x+2)}{LCD}-\frac{(x+3)^2}{LCD}=\frac{12x^2+5x-2-x^2-6x-9}{LCD}=\frac{11x^2-x-11}{LCD}$
Step 4. We conclude the results as $\frac{11x^2-x-11}{(2x-1)(x+3)(3x+2)}$ with $x\ne-3, -\frac{2}{3},\frac{1}{2}$.
