Solution set of $\dfrac{1}{x + 2} - \dfrac{1}{x + 5} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$ is—

Simplifying each side gives $\tfrac{3}{(x+2)(x+5)}=\tfrac{1}{(x+3)(x+4)}$, and cross-multiplying leads to $x^2+7x+13=0$. Its discriminant $49-52=-3<0$, so there is no real root and the solution set is empty.