jmc

Multiplying both sides by $x(x-2),$ we get $$2x^2+(2x^2-24)(x-2)=11x(x-2),$$which simplifies to $$2x^3-13x^2-2x+48=0.$$Looking for rational roots to the equation, we see that $x=6$ is a solution. Performing the polynomial division, we get $$2x^3-13x^2-2x+48=(x-6)(2x^2-x-8)=0,$$so either $x=6$ or $2x^2-x-8=0.$ The latter quadratic has solutions $$x=\frac{1\pm \sqrt{65}}{4},$$so the smallest root of the original equation is $x=\boxed{\frac{1-\sqrt{65}}{4}}.$