jmc

First, we simplify the denominator of the right side: $$2+\frac{1}{x-2}=\frac{2(x-2)}{x-2}+\frac{1}{x-2}=\frac{2x-4+1}{x-2}=\frac{2x-3}{x-2}.$$ So, we can now write the right side as $$\frac{1}{2+\frac{1}{x-2}}=\frac{1}{\frac{2x-3}{x-2}}=\frac{x-2}{2x-3},$$ so our equation is $$x=\frac{x-2}{2x-3}.$$ Multiplying both sides by $2x-3$ gives $$2x^2-3x=x-2,$$ so $2x^2-4x+2=0$. Dividing by 2 and factoring gives $(x-1{)}^2=0$, so the solution is $x=\boxed{1}$.