Question

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

Answer 1

а)

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

б)

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