Solve the following polynomial and quadratic inequalities 2 + 3/(x + 1) > 2/x

Rearranging the inequality 2 + 3/(x + 1) > 2/x gives 2 + (3x - 2x - 2) / (x(x + 1)) > 0, which simplifies to 2(x^2 + x + 1) / (x(x + 1)) > 0. Because x^2 + x + 1 is always positive for all real x, the denominator x(x + 1) must be positive, which occurs when x < -1 or x > 0. Solving the specific critical points in the options shows that intervals like x < -2, -1 < x < 0, and 1/2 < x all satisfy their respective sub-conditions within the broader valid solution sets, meaning all these ranges represent valid solutions. The result is that all of these are correct.

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