1. 计算：
（1）$\sqrt{27}-\sqrt{\dfrac{3}{2}}× \sqrt{8}$； （2）$\sqrt{12}+3\sqrt{1\dfrac{1}{3}}-\sqrt{27}$；
（3）$(\sqrt{24}-2\sqrt{\dfrac{3}{2}}+3\sqrt{2\dfrac{2}{3}})× \sqrt{3}$； （4）$(3\sqrt{18}+\dfrac{1}{5}\sqrt{50}-4\sqrt{\dfrac{1}{2}})÷ \sqrt{32}$；
（5）$\sqrt{12}× \sqrt{\dfrac{3}{2}}-\sqrt{10}÷ \sqrt{5}+\sqrt{8}$； （6）$\sqrt{\dfrac{2}{25}}-\sqrt{6}+2\sqrt{2}× \dfrac{\sqrt{2}}{4}÷ 5\sqrt{2}$.
（1）$\sqrt{27}-\sqrt{\dfrac{3}{2}}× \sqrt{8}$； （2）$\sqrt{12}+3\sqrt{1\dfrac{1}{3}}-\sqrt{27}$；
（3）$(\sqrt{24}-2\sqrt{\dfrac{3}{2}}+3\sqrt{2\dfrac{2}{3}})× \sqrt{3}$； （4）$(3\sqrt{18}+\dfrac{1}{5}\sqrt{50}-4\sqrt{\dfrac{1}{2}})÷ \sqrt{32}$；
（5）$\sqrt{12}× \sqrt{\dfrac{3}{2}}-\sqrt{10}÷ \sqrt{5}+\sqrt{8}$； （6）$\sqrt{\dfrac{2}{25}}-\sqrt{6}+2\sqrt{2}× \dfrac{\sqrt{2}}{4}÷ 5\sqrt{2}$.
答案：1.(1)$\sqrt{3}$ (2)$\sqrt{3}$ (3)$9\sqrt{2}$ (4)$2$ (5)$4\sqrt{2}$ (6)$\frac{3}{10}\sqrt{2}-\sqrt{6}$
解析：
解：原式$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+(2\sqrt{2}×\dfrac{\sqrt{2}}{4})÷5\sqrt{2}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+(\dfrac{4}{4})÷5\sqrt{2}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+\dfrac{1}{5\sqrt{2}}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+\dfrac{\sqrt{2}}{10}$
$=\dfrac{2\sqrt{2}}{10}+\dfrac{\sqrt{2}}{10}-\sqrt{6}$
$=\dfrac{3\sqrt{2}}{10}-\sqrt{6}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+(\dfrac{4}{4})÷5\sqrt{2}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+\dfrac{1}{5\sqrt{2}}$
$=\dfrac{\sqrt{2}}{5}-\sqrt{6}+\dfrac{\sqrt{2}}{10}$
$=\dfrac{2\sqrt{2}}{10}+\dfrac{\sqrt{2}}{10}-\sqrt{6}$
$=\dfrac{3\sqrt{2}}{10}-\sqrt{6}$
2. 计算：
（1）$(2+\sqrt{3})(2-\sqrt{3})+(\sqrt{2}-\sqrt{3})^{2}$； （2）$\dfrac{3-\sqrt{12}}{\sqrt{3}}+(3+\sqrt{3})(3-\sqrt{3})$；
（3）$\vert -\sqrt{2}\vert +(\sqrt{2}-\dfrac{1}{2})^{2}-(\sqrt{2}+\dfrac{1}{2})^{2}$； （4）$(2\sqrt{5}+3\sqrt{2})^{2}-(2\sqrt{5}-3\sqrt{2})^{2}$；
（5）$(4+\sqrt{15})^{2025}× (4-\sqrt{15})^{2026}$； （6）$(1+\sqrt{2}+\sqrt{3})× (1+\sqrt{2}-\sqrt{3})$.
（1）$(2+\sqrt{3})(2-\sqrt{3})+(\sqrt{2}-\sqrt{3})^{2}$； （2）$\dfrac{3-\sqrt{12}}{\sqrt{3}}+(3+\sqrt{3})(3-\sqrt{3})$；
（3）$\vert -\sqrt{2}\vert +(\sqrt{2}-\dfrac{1}{2})^{2}-(\sqrt{2}+\dfrac{1}{2})^{2}$； （4）$(2\sqrt{5}+3\sqrt{2})^{2}-(2\sqrt{5}-3\sqrt{2})^{2}$；
（5）$(4+\sqrt{15})^{2025}× (4-\sqrt{15})^{2026}$； （6）$(1+\sqrt{2}+\sqrt{3})× (1+\sqrt{2}-\sqrt{3})$.
答案：2.(1)$6 - 2\sqrt{6}$ (2)$\sqrt{3}+4$ (3)$-\sqrt{2}$ (4)$24\sqrt{10}$ (5)$4-\sqrt{15}$ (6)$2\sqrt{2}$
解析：
解：原式$=\dfrac{3}{\sqrt{3}}-\dfrac{\sqrt{12}}{\sqrt{3}}+(3+\sqrt{3})(3-\sqrt{3})$
$=\sqrt{3}-\sqrt{4}+(3^{2}-(\sqrt{3})^{2})$
$=\sqrt{3}-2+(9 - 3)$
$=\sqrt{3}-2 + 6$
$=\sqrt{3}+4$
$=\sqrt{3}-\sqrt{4}+(3^{2}-(\sqrt{3})^{2})$
$=\sqrt{3}-2+(9 - 3)$
$=\sqrt{3}-2 + 6$
$=\sqrt{3}+4$