5. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$
5. 5

Given $\because x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$,
$$
\therefore \quad x^{2}-8 x+13=0 \text {, }
$$

and the denominator of the fraction, when divided by $x^{2}-8 x+13$, leaves a remainder of 2, and the numerator of the fraction, when divided by $x^{2}-8 x+13$, leaves a remainder of 10.
$\therefore$ The value of the original expression is 5.