a) Ta có: \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
\(=\left(\sqrt{\frac{9}{4}\cdot6}+\sqrt{4\cdot\frac{2}{3}}-\sqrt{16\cdot\frac{3}{2}}\right)\left(\sqrt{9\cdot\frac{2}{3}}-2\sqrt{3}-\sqrt{6}\right)\)
\(=\left(\sqrt{\frac{27}{2}}+\sqrt{2}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-2\sqrt{3}-\sqrt{6}\right)\)
\(=-2\sqrt{3}\cdot\left(\sqrt{\frac{27}{2}}+\sqrt{2}-2\sqrt{6}\right)\)
\(=-\sqrt{12\cdot\frac{27}{2}}-2\sqrt{6}+4\sqrt{18}\)
\(=-9\sqrt{2}-2\sqrt{6}+12\sqrt{2}\)
\(=3\sqrt{2}-2\sqrt{6}\)
b) Ta có: \(\frac{4}{\sqrt{3}+1}-\frac{5}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}\)
\(=\frac{4\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\frac{5\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(\sqrt{3}+3\right)}{\left(\sqrt{3}-3\right)\left(\sqrt{3}+3\right)}\)
\(=\frac{4\left(\sqrt{3}-1\right)}{2}-\frac{5\left(\sqrt{3}+2\right)}{-1}+\frac{6\left(\sqrt{3}+3\right)}{-6}\)
\(=2\left(\sqrt{3}-1\right)+5\left(\sqrt{3}+2\right)-\left(\sqrt{3}+3\right)\)
\(=2\sqrt{3}-2+5\sqrt{3}+10-\sqrt{3}-3\)
\(=6\sqrt{3}+5\)