Meta-nested

$f=1+ \dfrac{\color{#3D99F6} 3+\dfrac{1+ \frac{3+\cdots}{2+\cdots}}{3+ \frac{1+\cdots}{3+\cdots}} } { \color{#D61F06} 2+\dfrac{2+\frac{2+\cdots}{1+\cdots}}{1+ \frac{3+\cdots}{2+\cdots}}} =\dfrac{A}{B}$

Define $\color{#3D99F6} g= 3+\dfrac{1+ \frac{3+\cdots}{2+\cdots}}{3+ \frac{1+\cdots}{3+\cdots}}$ and $\color{#D61F06} h= 2+\dfrac{2+\frac{2+\cdots}{1+\cdots}}{1+ \frac{3+\cdots}{2+\cdots}}$

We now have the equations:

$g=3+\dfrac{f}{g} \implies f=g(g-3)$

$h=2+\dfrac{h}{f} \implies \dfrac{1}{h}=\dfrac{1}{2}+\dfrac{1}{2f} \implies \dfrac{1}{h}=\dfrac{1}{2}+\dfrac{1}{2g(g-3)}$

$f=1+\dfrac{g}{h} \implies g(g-3)=1+g \left( \dfrac{1}{2}+\dfrac{1}{2g(g-3)} \right) \implies 2g^3-13g^2+19g+7=0$

This 3rd-degree polynomial on $g$ has three real solutions, but only one of them, $g=\frac{7}{2}$ , leads to a value of $f>1$ .

The solution is $f=\dfrac{7}{4}$ and the answer is $A=7, B=4$ and $\boxed{A+B=11}$ .

$\begin{aligned} f&=1+\dfrac{3+\dfrac{1+\dfrac{3+\cdots}{2+\cdots}}{3+\dfrac{1+\cdots}{3+\cdots}}}{2+\dfrac{2+\dfrac{2+\cdots}{1+\cdots}}{1+\dfrac{2+\cdots}{2+\cdots}}}=\dfrac{A}{B}\\ \text{Let,}\\ f&=1+\dfrac{x}{y} \hspace{4mm}\color{#3D99F6}\small(1) \\ \text{ where,}\\ x&=3+\dfrac{1+\dfrac{3+\cdots}{2+\cdots}}{3+\dfrac{1+\cdots}{3+\cdots}}\\\\ \text{and }y&=2+\dfrac{2+\dfrac{2+\cdots}{1+\cdots}}{1+\dfrac{3+\cdots}{2+\cdots}}\\ (y-2)&=\dfrac{y}{f}\\ \implies f(y-2)&=y\\ y(f-1)&=2f\\ y\cdot\dfrac{x}{y}&=2f \hspace{4mm}\color{#3D99F6}\small\text{From}(1)\\ \implies x&=2f\hspace{4mm}\color{#3D99F6}\small(2) \hspace{8mm},y>2\\\\ x&=3+\dfrac{1+\dfrac{3+\cdots}{2+\cdots}}{3+\dfrac{1+\cdots}{3+\cdots}}\\ x&=3+\dfrac{f}{x}\\ 2f&=3+\dfrac{f}{2f}\hspace{4mm}\color{#3D99F6}\small\text{From }(2)\\ 2f&=3+\dfrac{1}{2}\hspace{4mm}\color{#3D99F6}\small f>1\\ 2f&=\dfrac{7}{2}\\ f&=\dfrac{7}{4}\\ \dfrac{A}{B}&=\dfrac{7}{4}\\ \implies A+B&=\color{#EC7300}\boxed{\color{#333333}11} \end{aligned}$

1. $f=1+\frac{x}{y}$
2. $x=3+\frac{f}{x}$
3. $y=2+\frac{y}{f}$

Solving Eq. 2 for $x$ gives

$x^2-3x-f=0\\ x=\frac{3\pm \sqrt{9+4x}}{2}$ .

$x$ must be positive, and $\sqrt{9+4x}>3$ , so

$\bf x=\frac{3+\sqrt{9+4x}}{2}$ .

Solving Eq. 3 for $y$ gives

$\bf y=\frac{2}{1-\frac{1}{f}}$ .

Inserting these expressions into Eq. 1 and solving for $f$ goes as follows:

$f=1+\frac{\left(\frac{3+\sqrt{9+4f}}{2}\right)}{\left(\frac{2}{1-\frac{1}{f}}\right)}\\ 4(f-1)=(1-\frac{1}{f})(3+\sqrt{9+4f})\\ 4f(f-1)=(3+\sqrt{9+4f})(f-1)\\ \bf f=1\text{ or } 4f=3+\sqrt{9+4f}$

If $f=1$ , then $y$ is undefined, so $\bf 4f=3+\sqrt{9+4f}$ . Therefore,

$(4f-3)^2=9+4f\\ 16f^2-28f=0\\ \bf f=0\text{ or }f=\frac{7}{4}$ .

$f$ is clearly a positive number, so $f=\boxed{\bf \frac{7}{4}}$ . Since 7 and 4 are coprime, $A=7$ and $B=4$ , so $A+B=\boxed{\bf 11}$ .