12. Suppose $x=\frac{13}{\sqrt{19+8 \sqrt{3}}}$. Find the exact value of
$$
\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15} .
$$
12. Answer: 5 .
Note that $x=\frac{1}{13 \sqrt{(4+\sqrt{3})^{2}}}=\frac{13(4-\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}=4-\sqrt{3}$. So $(x-4)^{2}=3$. That is,
$$
x^{2}-8 x+15=2 \text {. }
$$

It follows that
$$
\begin{aligned}
\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15} & =x^{2}+2 x-1+\frac{38-20 x}{x^{2}-8 x+15} \\
& =x^{2}+2 x-1+\frac{38-20 x}{2} \\
& =x^{2}-8 x+18 \\
& =2+3=5 .
\end{aligned}
$$