Bài 6:
$1) A=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4}-\sqrt{15}\\A=(4+\sqrt{15})(\sqrt{10}-\sqrt{6}).2-\sqrt{15}\\A=(8+2\sqrt{15})(\sqrt{10}-\sqrt{6})-\sqrt{15}\\A=8\sqrt{10}-8\sqrt{6}+2\sqrt{150}-2\sqrt{90}-\sqrt{15}\\A=8\sqrt{10}-8\sqrt{6}+10\sqrt{6}-6\sqrt{10}-\sqrt{15}\\A=2\sqrt{10}+2\sqrt{6}-\sqrt{15}\\2)B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\\B=\dfrac{(3+2\sqrt{3})\sqrt{3}}{3}-(\sqrt{3}+\sqrt{2})+(2+\sqrt{2})(\sqrt{2}-1)\\B=\dfrac{3\sqrt{3}+6}{3}-\sqrt{3}-\sqrt{2}+2\sqrt{2}-2+2-\sqrt{2}\\B=\dfrac{3\sqrt{3}+6}{3}-\sqrt{3}+0\\B=\dfrac{3(\sqrt{3}+2)}{3}-\sqrt{3}\\B=\sqrt{3}+2-\sqrt{3}=2\\3)C=(\sqrt{6}-\sqrt{3}-1)(\sqrt{6}-\sqrt{3}+1)+6\sqrt{2}+\dfrac{8}{\sqrt{3}}+\dfrac{8-\sqrt{128}}{\sqrt{3}-2\sqrt{2}}\\=(\sqrt{6}-\sqrt{3})^{2}-1+6\sqrt{2}+\dfrac{8-8\sqrt{2}}{\sqrt{3}-2\sqrt{2}}\\=6-2\sqrt{18}+3-1+6\sqrt{2}+\dfrac{(8-8\sqrt{2})(\sqrt{3}+2\sqrt{2})}{5}\\=6-6\sqrt{2}+3-1+6\sqrt{2}+\dfrac{8\sqrt{3}+16\sqrt{2}-8\sqrt{6}-32}{-5}\\=8-\dfrac{8\sqrt{3}+16\sqrt{2}-8\sqrt{6}-32}{5}\\=\dfrac{40-(8\sqrt{3}+16\sqrt{2}-8\sqrt{6}-32)}{5}\\=\dfrac{40-8\sqrt{3}-16\sqrt{2}+8\sqrt{6}+32}{5}\\=\dfrac{72-8\sqrt{3}-16\sqrt{2}+8\sqrt{6}}{5}$
