Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}+\dfrac{1}{3}.\]
Combining the two fractions on the right side gives $\frac{2(3x+4)}{3(2x+9)}+\dfrac{1}{3} = \frac{6x+8}{6x+27}+\frac{1}{3}$. Combining over a common denominator gives $\frac{6x+8}{6x+27}+\frac{1}{3} = \frac{6x+8}{6x+27}+\frac{1}{3} \cdot \frac{9}{9} = \frac{6x+8}{6x+27}+\frac{9}{27} = \frac{6x+8}{6x+27}+\frac{1}{3}$ $= \frac{6x+8+2(6x+27)}{9 \cdot 3}=\frac{14x+62}{9 \cdot 3} = \frac{14x+62}{27}$.
Our inequality becomes $\frac{x+1}{x+2}>\frac{14x+62}{27}$, and we get rid of the fraction by multiplying both sides by $27(x+2)$, $27(x+2) \cdot \frac{x+1}{x+2} > 27(x+2) \cdot \frac{14x+62}{27}$. This gives $[27(x+2)] \frac{x+1}{\cancel{x+2}} > [27(x+2)] \frac{14x+62}{\cancel{27}}$ $\Rightarrow 27(x+1) > (x+2)(14x+62)$ $\Rightarrow27x+27 > (x+2)(14x+62)$.
Expanding the right side gives $27x+27 > 14x^2+78x+124x+124$, which simplifies to $27x+27 > 14x^2+202x+124$ and $0 > 14x^2+175x+97$. Trying to factor this quadratic limited by the Rational Root Theorem, we find that $x = 1$ does not work but $x = -\frac{97}{14}$ does. Therefore, since $\frac{x+1}{x+2}$ is defined when $x \neq -2$, and after testing we find that $x > -\frac{97}{14}$ for other real numbers $x$, our answer is all $x$ such that $x \in \boxed{\left(-\frac{97}{14}, -2\right)}$.
To solve the inequality, we need to simplify the expression on both sides and then isolate the variable. Let's start by simplifying the expression:
\[\frac{x+1}{x+2} > \frac{3x+4}{2x+9} + \frac{1}{3}.\]
First, we need to find a common denominator for the two fractions on the right side. The common denominator is given by \(3(2x+9)\):
\[\frac{x+1}{x+2} > \frac{3x+4}{2x+9}\cdot\frac{3}{3} + \frac{1}{3}.\]
Simplifying the expression on the right side:
\[\frac{x+1}{x+2} > \frac{9(3x+4)}{3(2x+9)} + \frac{2x+9}{3(2x+9)}.\]
Now, we can combine the two fractions on the right side:
\[\frac{x+1}{x+2} > \frac{9(3x+4) + 2x+9}{3(2x+9)}.\]
Expanding and simplifying the numerator:
\[\frac{x+1}{x+2} > \frac{27x + 36 + 2x + 9}{3(2x+9)}.\]
Combining like terms in the numerator:
\[\frac{x+1}{x+2} > \frac{29x + 45}{3(2x+9)}.\]
Now, let's eliminate the denominator by multiplying both sides of the inequality by \(3(2x+9)\):
\[3(2x+9)\cdot\frac{x+1}{x+2} > (29x + 45).\]
Expanding the numerator:
\[(2x+9)(x+1) > (29x + 45).\]
Expanding and simplifying the left side:
\[2x^2 + 11x + 9 > 29x + 45.\]
Moving all terms to one side:
\[2x^2 + 11x - 29x + 9 - 45 > 0.\]
Combining like terms:
\[2x^2 - 18x - 36 > 0.\]
Now, we have a quadratic inequality. To solve it, we can factor the quadratic expression:
\[2(x^2 - 9x - 18) > 0.\]
Factoring:
\[2(x-6)(x+3) > 0.\]
Now, we can determine the sign of each factor to find the intervals that satisfy the inequality. We have three critical points: -3, 0, and 6.
For \(x < -3\), both factors are negative, so the inequality is not satisfied.
For \(-3 < x < 0\), the first factor is positive, and the second factor is negative, so the inequality is satisfied.
For \(0 < x < 6\), both factors are positive, so the inequality is satisfied.
For \(x > 6\), both factors are positive, so the inequality is satisfied.
Finally, we can write the solution as an interval:
\((-3, 0) \cup (0, 6) \cup (6, +\infty).\)
Therefore, the solution to the inequality is \(\boxed{(-3, 0) \cup (0, 6) \cup (6, +\infty)}.\)