e) \( \frac{2}{x^{2}+3 x+2}-\frac{1}{x^{2}+x-2}-\frac{1}{x^{2}-1} \)

Let’s start by factoring the denominators in the expression: - \(x^{2}+3x+2 = (x+1)(x+2)\), - \(x^{2}+x-2 = (x-1)(x+2)\), - \(x^{2}-1 = (x-1)(x+1)\). Now we can rewrite the expression: \[ \frac{2}{(x+1)(x+2)} - \frac{1}{(x-1)(x+2)} - \frac{1}{(x-1)(x+1)} \] The common denominator for all three fractions will be \((x+1)(x+2)(x-1)\). Now, rewriting each term with this common denominator, we have: \[ \frac{2(x-1)}{(x+1)(x+2)(x-1)} - \frac{1(x+1)}{(x-1)(x+2)(x+1)} - \frac{1(x+2)}{(x-1)(x+1)(x+2)} \] Simplifying each of these fractions gives: \[ \frac{2(x-1)}{(x+1)(x+2)(x-1)} - \frac{(x+1)}{(x-1)(x+2)(x+1)} - \frac{(x+2)}{(x-1)(x+1)(x+2)} \] Now combine the numerators over the common denominator: \( = \frac{2(x-1) - (x+1) - (x+2)}{(x+1)(x+2)(x-1)} \) Combining the terms in the numerator: \[ = \frac{2x - 2 - x - 1 - x - 2}{(x+1)(x+2)(x-1)} \] \[ = \frac{-5}{(x+1)(x+2)(x-1)} \] So the final expression simplified is: \[ \frac{-5}{(x+1)(x+2)(x-1)} \]