Question

2 и 3 задачи для умных

Answer 1

Надеюсь все разберете

См. Фото

Answer 2

2 правильно, но 4 нет

Answer 3

$2) sqrt{31-6sqrt{26}}-sqrt{31+6sqrt{26}}=\ \=sqrt{31-2*3sqrt{2}sqrt{13}}-sqrt{31+2*3sqrt{2}sqrt{13}}=\\=sqrt{(3sqrt{2})^2-2*3sqrt{2}sqrt{13}+(sqrt{13})^2}-sqrt{(3sqrt{2})^2+2*3sqrt{2}sqrt{13}+(sqrt{13})^2}=\\=sqrt{(3sqrt{2}-sqrt{13})^2}-sqrt{(3sqrt{2}+sqrt{13})^2}=\\=3sqrt{2}-sqrt{13}-(3sqrt{2}+sqrt{13})=\\=3sqrt{2}-sqrt{13}-3sqrt{2}-sqrt{13}=-2sqrt{13}$

$4)sqrt[3]{5sqrt{2}-7}*sqrt{3+2sqrt{2}}=\ \=sqrt[3*2]{(5sqrt{2}-7)^2}*sqrt[2*3]{(3+2sqrt{2})^3}=\\=sqrt[6]{(5sqrt{2})^2-2*5sqrt{2}*7+7^2}*sqrt[6]{3^3+3*3^2*2sqrt{2}+3*3(2sqrt{2})^2+(2sqrt{2})^3}=\\=sqrt[6]{50-70sqrt{2}+49}*sqrt[6]{27+54sqrt{2}+72+16sqrt{2}}=\\=sqrt[6]{99-70sqrt{2}}*sqrt[6]{99+70sqrt{2}}=\\=sqrt[6]{(99-70sqrt{2})(99+70sqrt{2})}=\\=sqrt[6]{99^2-(70sqrt{2})^2}=sqrt[6]{9801-9800}=sqrt[6]{1}=1$