15. (本小题 8 分)计算.
(1) $(\sqrt{48}+\dfrac{1}{2}\sqrt{1\dfrac{1}{2}})×\dfrac{\sqrt{3}}{3}$;
(2) $(4\sqrt{54}+\sqrt{50}-\dfrac{\sqrt{6}}{2})÷\sqrt{2}$.

(1)
解：
首先化简各项：
$\sqrt{48} = \sqrt{16 × 3} = 4\sqrt{3}$，
$\dfrac{1}{2}\sqrt{1\dfrac{1}{2}} = \dfrac{1}{2}\sqrt{\dfrac{3}{2}} = \dfrac{1}{2} × \dfrac{\sqrt{6}}{2} = \dfrac{\sqrt{6}}{4}$，
将化简后的项代入原式并乘以$\dfrac{\sqrt{3}}{3}$：
$(\sqrt{48}+\dfrac{1}{2}\sqrt{1\dfrac{1}{2}}) × \dfrac{\sqrt{3}}{3} = (4\sqrt{3} + \dfrac{\sqrt{6}}{4}) × \dfrac{\sqrt{3}}{3}$
$= 4\sqrt{3} × \dfrac{\sqrt{3}}{3} + \dfrac{\sqrt{6}}{4} × \dfrac{\sqrt{3}}{3}$
$= 4 + \dfrac{\sqrt{2}}{4}$
(2)
解：
首先化简各项：
$4\sqrt{54} = 4\sqrt{9 × 6} = 12\sqrt{6}$，
$\sqrt{50} = \sqrt{25 × 2} = 5\sqrt{2}$，
将化简后的项代入原式并除以$\sqrt{2}$：
$(4\sqrt{54} + \sqrt{50} - \dfrac{\sqrt{6}}{2}) ÷ \sqrt{2} = (12\sqrt{6} + 5\sqrt{2} - \dfrac{\sqrt{6}}{2}) ÷ \sqrt{2}$
$= 12\sqrt{6} ÷ \sqrt{2} + 5\sqrt{2} ÷ \sqrt{2} - \dfrac{\sqrt{6}}{2} ÷ \sqrt{2}$
$= 12\sqrt{3} + 5 - \dfrac{\sqrt{3}}{2}$
$= \dfrac{23\sqrt{3}}{2} + 5（或 5+\dfrac{23\sqrt{3}}{2}）$

16. (本小题 8 分)计算.
(1) $\sqrt{(-3)^{2}}+2×(\sqrt{2}-1)-\vert-2\sqrt{2}\vert$;
(2) $(\sqrt{3}-\sqrt{2})^{2}(2\sqrt{6}+5)$.

(1) $\sqrt{(-3)^{2}}+2×(\sqrt{2}-1)-\vert-2\sqrt{2}\vert$
$=3 + 2\sqrt{2}-2 - 2\sqrt{2}$
$=(3 - 2)+(2\sqrt{2}-2\sqrt{2})$
$=1 + 0$
$=1$
(2) $(\sqrt{3}-\sqrt{2})^{2}(2\sqrt{6}+5)$
$=(3 - 2\sqrt{6}+2)(2\sqrt{6}+5)$
$=(5 - 2\sqrt{6})(5 + 2\sqrt{6})$
$=5^{2}-(2\sqrt{6})^{2}$
$=25 - 24$
$=1$