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Автор - zamira4ka05

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Ответ

Проверено экспертом

Автор - Universalka

1)sqrt{6+2sqrt{5}}-sqrt{5}=sqrt{5+2sqrt{5}+1}-sqrt{5}=sqrt{(sqrt{5}+1)^{2}}-sqrt{5  }=|sqrt{5}+1|-sqrt{5}=sqrt{5}+1-sqrt{5}=1\\Otvet:boxed{1}\\2)sqrt{3}-sqrt{7+4sqrt{3}}=sqrt{3}-sqrt{4+4sqrt{3}+3}=sqrt{3}-sqrt{(2+sqrt{3})^{2}}=sqrt{3}-|2+sqrt{3}|=sqrt{3}-(2+sqrt{3})=sqrt{3}-2-sqrt{3}=-2\\Otvet:boxed{-2}

sqrt{9-2sqrt{14}}+sqrt{9+2sqrt{14}}=sqrt{7-2sqrt{14}+2}+sqrt{7+2sqrt{14}+2}=sqrt{(sqrt{7}-sqrt{2})^{2}}+sqrt{(sqrt{7}+sqrt{2})^{2}}=|sqrt{7}-sqrt{2}|+|sqrt{7}+sqrt{2}|=sqrt{7}-sqrt{2}+sqrt{7}+sqrt{2}=2sqrt{7}\\Otvet:boxed{2sqrt{7}}

4)sqrt{23+8sqrt{7}}-sqrt{23-8sqrt{7}}=sqrt{16+8sqrt{7}+7}-sqrt{16-8sqrt{7}+7}=sqrt{(4+sqrt{7})^{2}}-sqrt{(4-sqrt{7})^{2}}=|4+sqrt{7}|-|4-sqrt{7}|=(4+sqrt{7})-(4-sqrt{7})=4+sqrt{7}-4+sqrt{7}=2sqrt{7}\\Otvet:boxed{2sqrt{7}}

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