làm từ câu 1 đến câu 7 ở trong hình

Bài 6: 1) $\frac{4-4\sqrt5}{\sqrt2-\sqrt{10}}=\frac{4(1-\sqrt5)}{\sqrt2(1-\sqrt5)}=\frac{4}{\sqrt2}=\frac{4\sqrt2}{2}=2\sqrt2$ 2) $\frac{2\sqrt2-4\sqrt3}{\sqrt{35}-6\sqrt{15}}=\frac{2(\sqrt2-2\sqrt3)}{\sqrt5(\sqrt7-6\sqrt3)}=\frac{2(\sqrt2-2\sqrt3)\times (\sqrt7+6\sqrt3)}{\sqrt5\times (\sqrt7-6\sqrt3)\times (\sqrt7+6\sqrt3)}=\frac{2(\sqrt{14}+12\sqrt6-2\sqrt{21}-12\sqrt9)}{\sqrt5\times (7-108)}=\frac{2(\sqrt{14}+12\sqrt6-2\sqrt{21}-36)}{-101\sqrt5}=\frac{2(36-\sqrt{14}-12\sqrt6+2\sqrt{21})}{101\sqrt5}=\frac{2(36-\sqrt{14}-12\sqrt6+2\sqrt{21})\times \sqrt5}{101\times 5}=\frac{2(36\sqrt5-\sqrt{70}-12\sqrt{30}+2\sqrt{105})}{505}$ 3) $\frac{2\sqrt2}{\sqrt{40}}=\frac{2\sqrt2}{2\sqrt{10}}=\frac{\sqrt2}{\sqrt{10}}=\frac{\sqrt2\times \sqrt{10}}{10}=\frac{2\sqrt5}{10}=\frac{\sqrt5}{5}$ 4) $\frac{\sqrt6-\sqrt{21}}{\sqrt2-\sqrt7}-\frac2{3\sqrt3+5}=\frac{\sqrt3(\sqrt2-\sqrt7)}{\sqrt2-\sqrt7}-\frac{2\times (3\sqrt3-5)}{(3\sqrt3+5)\times (3\sqrt3-5)}=\sqrt3-\frac{6\sqrt3-10}{27-25}=\sqrt3-\frac{6\sqrt3-10}{2}=\frac{2\sqrt3-(6\sqrt3-10)}{2}=\frac{2\sqrt3-6\sqrt3+10}{2}=\frac{10-4\sqrt3}{2}=5-2\sqrt3$ 5) $\frac{3\sqrt2-\sqrt6}{1-\sqrt3}-\frac5{1-\sqrt6}=\frac{\sqrt6(1-\sqrt3)}{1-\sqrt3}-\frac{5\times (1+\sqrt6)}{(1-\sqrt6)\times (1+\sqrt6)}=\sqrt6-\frac{5+5\sqrt6}{1-6}=\sqrt6-\frac{5+5\sqrt6}{-5}=\sqrt6+\frac{5+5\sqrt6}{5}=\sqrt6+1+\sqrt6=1+2\sqrt6$ 6) $\frac4{1+\sqrt5}-\frac{\sqrt{20}-5}{2-\sqrt5}=\frac{4\times (1-\sqrt5)}{(1+\sqrt5)\times (1-\sqrt5)}-\frac{2\sqrt5-5}{2-\sqrt5}=\frac{4-4\sqrt5}{1-5}-\frac{2\sqrt5-5}{2-\sqrt5}=\frac{4-4\sqrt5}{-4}-\frac{(2\sqrt5-5)\times (2+\sqrt5)}{(2-\sqrt5)\times (2+\sqrt5)}=-1+\sqrt5-\frac{4\sqrt5+10-10-5\sqrt5}{4-5}=-1+\sqrt5-\frac{-\sqrt5}{-1}=-1+\sqrt5-\sqrt5=-1$ 7) $\frac{3\sqrt2+2\sqrt3}{\sqrt2+\sqrt3}-\frac2{\sqrt6-2}=\frac{\sqrt2(3+\sqrt6)}{\sqrt2+\sqrt3}-\frac{2\times (\sqrt6+2)}{(\sqrt6-2)\times (\sqrt6+2)}=\frac{\sqrt2(3+\sqrt6)}{\sqrt2+\sqrt3}-\frac{2\times (\sqrt6+2)}{6-4}=\frac{\sqrt2(3+\sqrt6)}{\sqrt2+\sqrt3}-\frac{2\times (\sqrt6+2)}{2}=\frac{\sqrt2(3+\sqrt6)}{\sqrt2+\sqrt3}-\sqrt6-2=\frac{\sqrt2(3+\sqrt6)-(\sqrt6+2)\times (\sqrt2+\sqrt3)}{\sqrt2+\sqrt3}=\frac{3\sqrt2+2\sqrt3-(\sqrt{12}+2\sqrt2+3\sqrt2+2\sqrt3)}{\sqrt2+\sqrt3}=\frac{3\sqrt2+2\sqrt3-(2\sqrt3+5\sqrt2)}{\sqrt2+\sqrt3}=\frac{3\sqrt2+2\sqrt3-2\sqrt3-5\sqrt2}{\sqrt2+\sqrt3}=\frac{-2\sqrt2}{\sqrt2+\sqrt3}=\frac{-2\sqrt2\times (\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)\times (\sqrt2-\sqrt3)}=\frac{-2\sqrt2\times (\sqrt2-\sqrt3)}{2-3}=\frac{-2\sqrt2\times (\sqrt2-\sqrt3)}{-1}=2\sqrt2\times (\sqrt2-\sqrt3)=4-2\sqrt6$ 8) $\frac5{\sqrt7-\sqrt2}-\frac{\sqrt{10}-\sqrt2}{\sqrt5-1}=\frac{5\times (\sqrt7+\sqrt2)}{(\sqrt7-\sqrt2)\times (\sqrt7+\sqrt2)}-\frac{\sqrt2(\sqrt5-1)}{\sqrt5-1}=\frac{5\times (\sqrt7+\sqrt2)}{7-2}-\sqrt2=\frac{5\times (\sqrt7+\sqrt2)}{5}-\sqrt2=\sqrt7+\sqrt2-\sqrt2=\sqrt7$ 9) $\frac{2\sqrt{24}+6}{\sqrt3+2\sqrt2}-\frac{10}{\sqrt8-\sqrt3}=\frac{4\sqrt6+6}{\sqrt3+2\sqrt2}-\frac{10\times (\sqrt8+\sqrt3)}{(\sqrt8-\sqrt3)\times (\sqrt8+\sqrt3)}=\frac{2(2\sqrt6+3)}{\sqrt3+2\sqrt2}-\frac{10\times (\sqrt8+\sqrt3)}{8-3}=\frac{2(2\sqrt6+3)}{\sqrt3+2\sqrt2}-\frac{10\times (\sqrt8+\sqrt3)}{5}=\frac{2(2\sqrt6+3)}{\sqrt3+2\sqrt2}-2\times (\sqrt8+\sqrt3)=\frac{2(2\sqrt6+3)-2\times (\sqrt8+\sqrt3)\times (\sqrt3+2\sqrt2)}{\sqrt3+2\sqrt2}=\frac{4\sqrt6+6-2\times (\sqrt{24}+4\sqrt6+3+2\sqrt6)}{\sqrt3+2\sqrt2}=\frac{4\sqrt6+6-2\times (4\sqrt6+3+6\sqrt6)}{\sqrt3+2\sqrt2}=\frac{4\sqrt6+6-2\times (10\sqrt6+3)}{\sqrt3+2\sqrt2}=\frac{4\sqrt6+6-20\sqrt6-6}{\sqrt3+2\sqrt2}=\frac{-16\sqrt6}{\sqrt3+2\sqrt2}=\frac{-16\sqrt6\times (\sqrt3-2\sqrt2)}{(\sqrt3+2\sqrt2)\times (\sqrt3-2\sqrt2)}=\frac{-16\sqrt6\times (\sqrt3-2\sqrt2)}{3-8}=\frac{-16\sqrt6\times (\sqrt3-2\sqrt2)}{-5}=\frac{16\sqrt6\times (\sqrt3-2\sqrt2)}{5}=\frac{16\times (3\sqrt2-4\sqrt3)}{5}=\frac{48\sqrt2-64\sqrt3}{5}$ 10) $\frac{3\sqrt2-\sqrt{10}}{3-\sqrt5}-\frac2{1+\sqrt2}=\frac{\sqrt2(3-\sqrt5)}{3-\sqrt5}-\frac{2\times (1-\sqrt2)}{(1+\sqrt2)\times (1-\sqrt2)}=\sqrt2-\frac{2\times (1-\sqrt2)}{1-2}=\sqrt2-\frac{2\times (1-\sqrt2)}{-1}=\sqrt2+2\times (1-\sqrt2)=\sqrt2+2-2\sqrt2=2-\sqrt2$ 11) $\frac{3\sqrt2+2\sqrt3}{\sqrt6}+\frac1{\sqrt3+\sqrt2}=\frac{\sqrt6(3+\sqrt2)}{\sqrt6}+\frac{1\times (\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)\times (\sqrt3-\sqrt2)}=3+\sqrt2+\frac{\sqrt3-\sqrt2}{3-2}=3+\sqrt2+\sqrt3-\sqrt2=3+\sqrt3$ 12) $\frac{\sqrt{21}-7\sqrt7}{\sqrt{3-7}}+\frac{18}{\sqrt7+5}=\frac{\sqrt7(\sqrt3-7)}{\sqrt{3-7}}+\frac{18\times (\sqrt7-5)}{(\sqrt7+5)\times (\sqrt7-5)}=\frac{\sqrt7(\sqrt3-7)}{\sqrt{3-7}}+\frac{18\times (\sqrt7-5)}{7-25}=\frac{\sqrt7(\sqrt3-7)}{\sqrt{3-7}}+\frac{18\times (\sqrt7-5)}{-18}=\frac{\sqrt7(\sqrt3-7)}{\sqrt{3-7}}-\sqrt7+5=\frac{\sqrt7(\sqrt3-7)-(\sqrt7+5)\times (\sqrt{3-7})}{\sqrt{3-7}}=\frac{\sqrt7(\sqrt3-7)-\sqrt{21-49}-5\sqrt{3-7}}{\sqrt{3-7}}=\frac{\sqrt7(\sqrt3-7)-\sqrt{-28}-5\sqrt{3-7}}{\sqrt{3-7}}=\frac{\sqrt7(\sqrt3-7)-2i\sqrt7-5\sqrt{3-7}}{\sqrt{3-7}}=\frac{\sqrt7(\sqrt3-7-2i-5\sqrt{3-7})}{\sqrt{3-7}}=\sqrt7(\sqrt3-7-2i-5\sqrt{3-7})$ 13) $\frac{2\sqrt3-\sqrt6}{\sqrt2-1}-\frac5{\sqrt6+1}=\frac{\sqrt3(2-\sqrt2)}{\sqrt2-1}-\frac{5\times (\sqrt6-1)}{(\sqrt6+1)\times (\sqrt6-1)}=\frac{\sqrt3(2-\sqrt2)}{\sqrt2-1}-\frac{5\times (\sqrt6-1)}{6-1}=\frac{\sqrt3(2-\sqrt2)}{\sqrt2-1}-\frac{5\times (\sqrt6-1)}{5}=\frac{\sqrt3(2-\sqrt2)}{\sqrt2-1}-\sqrt6+1=\frac{\sqrt3(2-\sqrt2)-(\sqrt6+1)\times (\sqrt2-1)}{\sqrt2-1}=\frac{2\sqrt3-\sqrt6-(\sqrt{12}+2\sqrt3-\sqrt6-1)}{\sqrt2-1}=\frac{2\sqrt3-\sqrt6-(2\sqrt3-\sqrt6-1)}{\sqrt2-1}=\frac{2\sqrt3-\sqrt6-2\sqrt3+\sqrt6+1}{\sqrt2-1}=\frac{1}{\sqrt2-1}=\frac{1\times (\sqrt2+1)}{(\sqrt2-1)\times (\sqrt2+1)}=\frac{\sqrt2+1}{2-1}=\sqrt2+1$ 14) $\frac3{\sqrt3-2}-\frac{12}{3+\sqrt3}=\frac{3\times (\sqrt3+2)}{(\sqrt3-2)\times (\sqrt3+2)}-\frac{12\times (3-\sqrt3)}{(3+\sqrt3)\times (3-\sqrt3)}=\frac{3\times (\sqrt3+2)}{3-4}-\frac{12\times (3-\sqrt3)}{9-3}=\frac{3\times (\sqrt3+2)}{-1}-\frac{12\times (3-\sqrt3)}{6}=-3\times (\sqrt3+2)-2\times (3-\sqrt3)=-3\sqrt3-6-6+2\sqrt3=-\sqrt3-12$ 15) $\frac1{2-\sqrt5}-\frac{\sqrt{10}+\sqrt5}{\sqrt2+1}=\frac{1\times (2+\sqrt5)}{(2-\sqrt5)\times (2+\sqrt5)}-\frac{\sqrt5(\sqrt2+1)}{\sqrt2+1}=\frac{2+\sqrt5}{4-5}-\sqrt5=\frac{2+\sqrt5}{-1}-\sqrt5=-2-\sqrt5-\sqrt5=-2-2\sqrt5$