16. (本小题 6 分)计算:$(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) - (\sqrt{3} + \sqrt{5})^2$.

解：原式$=(\sqrt{5})^2 - (\sqrt{3})^2 - [(\sqrt{3})^2 + 2\sqrt{3}×\sqrt{5} + (\sqrt{5})^2]$
$=5 - 3 - (3 + 2\sqrt{15} + 5)$
$=2 - (8 + 2\sqrt{15})$
$=2 - 8 - 2\sqrt{15}$
$=-6 - 2\sqrt{15}$

17. (本小题 6 分)计算:$\sqrt{12} ÷ \sqrt{3} + |3 - \sqrt{12}| - \frac{6}{\sqrt{3}}$.

$-1$

$\begin{aligned}&\sqrt{12} ÷ \sqrt{3} + |3 - \sqrt{12}| - \frac{6}{\sqrt{3}}\\=&\sqrt{\frac{12}{3}} + |3 - 2\sqrt{3}| - \frac{6\sqrt{3}}{\sqrt{3}×\sqrt{3}}\\=&\sqrt{4} + (2\sqrt{3} - 3) - \frac{6\sqrt{3}}{3}\\=&2 + 2\sqrt{3} - 3 - 2\sqrt{3}\\=&(2 - 3) + (2\sqrt{3} - 2\sqrt{3})\\=&-1 + 0\\=&-1\end{aligned}$

18. (本小题 6 分)已知$a = 3 + 2\sqrt{2},b = 3 - 2\sqrt{2}$,求$a^2b - ab^2$的值.

$4\sqrt{2}$

解：
因为 $a = 3 + 2\sqrt{2}$，$b = 3 - 2\sqrt{2}$，
所以 $ab = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$，
$a - b = (3 + 2\sqrt{2}) - (3 - 2\sqrt{2}) = 4\sqrt{2}$，
则 $a^2b - ab^2 = ab(a - b) = 1 × 4\sqrt{2} = 4\sqrt{2}$。