jmc

We can tell that $x-1=\frac{\sqrt{2}}{1+\frac{\sqrt{2}}{1+...}}$, and then $\frac{\sqrt{2}}{x-1}=1+\frac{\sqrt{2}}{1+\frac{\sqrt{2}}{1+...}}=x$. Solving for $x$, we find $\sqrt{2}=x(x-1)$, which means $x^2-x=\sqrt{2}$. Simplify the denominator of $\frac{1}{(x+1)(x-2)}$ to obtain $\frac{1}{x^2-x-2}$. Substituting for $x^2-x$, we get $\frac{1}{(x+1)(x-2)}=\frac{1}{\sqrt{2}-2}$. To rationalize the denominator, we multiply by the conjugate of $\sqrt{2}-2$. We have $\frac{1}{\sqrt{2}-2}=\frac{1\cdot (\sqrt{2}+2)}{(\sqrt{2}-2)\cdot (\sqrt{2}+2)}=\frac{\sqrt{2}+2}{2-4}=\frac{2+\sqrt{2}}{-2}.$ Here, we have $A=2,B=2$, and $C=-2$. So, taking the sum of the absolute values of $A$, $B$, and $C$ yields $\boxed{6}$.