Question

Упражнение 22 а и б решите пожалуйста, а то я болел, и пропустил тему.

Answer 1

а)

$frac{2-2sqrt{2}+sqrt{3}-sqrt{5}-sqrt{6}+sqrt{10}}{2+sqrt{3}-sqrt{5}} = frac{2+4sqrt{3} -2sqrt{2}-4sqrt{6}}{2(1+2sqrt{3}} = frac{2(1+2sqrt{3}-sqrt{2}-2sqrt{6})}{2(1+2sqrt{3})}=\\=frac{1+2sqrt{3}-sqrt{2}-2sqrt{6}}{1+2sqrt{3}} = frac{(1+2sqrt{3}-sqrt{2}-2sqrt{6})(1-2sqrt{3})}{-11} = frac{1-12-sqrt{2}+12sqrt{2}}{-11}=\\=frac{-11+11sqrt{2}}{-11} = frac{11(-1+sqrt{2})}{-11} = - (-1+sqrt{2}) = 1 -sqrt{2}$

б)

$frac{(sqrt{2}+sqrt{3})(2-sqrt{6}+sqrt{7})-sqrt{21}}{sqrt{7}-1} = frac{2sqrt{2}-sqrt{12}+sqrt{14} + 2sqrt{3}-sqrt{18}+sqrt{21}-sqrt{21}}{sqrt{7}-1}=\\=frac{2sqrt{2}-2sqrt{3}+sqrt{14}+2sqrt{3}-sqrt{18}}{sqrt{7}-1}=frac{2sqrt{2}+sqrt{14}-3sqrt{2}}{sqrt{7}-1}=frac{-sqrt{2}+sqrt{14}}{sqrt{7}-1}=\\=frac{(-sqrt{2}+sqrt{14}(sqrt{7}+1)}{(sqrt{7}-1)(sqrt{7}+1)} = frac{sqrt{14}-sqrt{2}+sqrt{98}+sqrt{14}}{7-1} = frac{-sqrt{2}+7sqrt{2}}{6}=frac{6sqrt{2}}{6}=sqrt{2}$

Answer 2

Спасибо огромное

Answer 3

не за что :)