Find the set of solutions for $x$ in the inequality $\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}$ when $x\neq 2$, $x\neq -\dfrac{9}{2}$.
1. Start with the given inequality:
   \[
   \frac{x+1}{x+2} > \frac{3x+4}{2x+9}
   \]
   We need to bring everything to one side to compare the expressions. Subtract \(\frac{3x+4}{2x+9}\) from both sides:
   \[
   \frac{x+1}{x+2} - \frac{3x+4}{2x+9} > 0
   \]

2. To combine the fractions, find a common denominator, which is \((x+2)(2x+9)\):
   \[
   \frac{(x+1)(2x+9) - (3x+4)(x+2)}{(x+2)(2x+9)} > 0
   \]

3. Expand the numerators:
   \[
   (x+1)(2x+9) = 2x^2 + 9x + 2x + 9 = 2x^2 + 11x + 9
   \]
   \[
   (3x+4)(x+2) = 3x^2 + 6x + 4x + 8 = 3x^2 + 10x + 8
   \]

4. Subtract the second expanded numerator from the first:
   \[
   2x^2 + 11x + 9 - (3x^2 + 10x + 8) = 2x^2 + 11x + 9 - 3x^2 - 10x - 8 = -x^2 + x + 1
   \]

5. The inequality now becomes:
   \[
   \frac{-x^2 + x + 1}{(x+2)(2x+9)} > 0
   \]

6. Factor the numerator \(-x^2 + x + 1\):
   \[
   -x^2 + x + 1 = -(x^2 - x - 1) = -(x - \frac{1+\sqrt{5}}{2})(x - \frac{1-\sqrt{5}}{2})
   \]

7. The inequality is now:
   \[
   \frac{-(x - \frac{1+\sqrt{5}}{2})(x - \frac{1-\sqrt{5}}{2})}{(x+2)(2x+9)} > 0
   \]

8. Identify the critical points where the expression changes sign:
   \[
   x = \frac{1+\sqrt{5}}{2}, \quad x = \frac{1-\sqrt{5}}{2}, \quad x = -2, \quad x = -\frac{9}{2}
   \]

9. Determine the sign of the expression in each interval defined by these critical points:
   - For \(x \in (-\infty, -\frac{9}{2})\):
     \[
     (x - \frac{1+\sqrt{5}}{2}) > 0, \quad (x - \frac{1-\sqrt{5}}{2}) > 0, \quad (x+2) < 0, \quad (2x+9) < 0
     \]
     The product of these signs is positive.

   - For \(x \in (-\frac{9}{2}, -2)\):
     \[
     (x - \frac{1+\sqrt{5}}{2}) > 0, \quad (x - \frac{1-\sqrt{5}}{2}) > 0, \quad (x+2) < 0, \quad (2x+9) > 0
     \]
     The product of these signs is negative.

   - For \(x \in (-2, \frac{1-\sqrt{5}}{2})\):
     \[
     (x - \frac{1+\sqrt{5}}{2}) > 0, \quad (x - \frac{1-\sqrt{5}}{2}) < 0, \quad (x+2) > 0, \quad (2x+9) > 0
     \]
     The product of these signs is positive.

   - For \(x \in (\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})\):
     \[
     (x - \frac{1+\sqrt{5}}{2}) < 0, \quad (x - \frac{1-\sqrt{5}}{2}) < 0, \quad (x+2) > 0, \quad (2x+9) > 0
     \]
     The product of these signs is negative.

   - For \(x \in (\frac{1+\sqrt{5}}{2}, \infty)\):
     \[
     (x - \frac{1+\sqrt{5}}{2}) > 0, \quad (x - \frac{1-\sqrt{5}}{2}) > 0, \quad (x+2) > 0, \quad (2x+9) > 0
     \]
     The product of these signs is positive.

10. The inequality \(\frac{-(x - \frac{1+\sqrt{5}}{2})(x - \frac{1-\sqrt{5}}{2})}{(x+2)(2x+9)} > 0\) is satisfied in the intervals where the product is negative:
    \[
    x \in (-\frac{9}{2}, -2) \cup \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)
    \]

The final answer is \( \boxed{ x \in (-\frac{9}{2}, -2) \cup \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) } \)