Question

Exercitiul 10 b si c.Raspunsiri serioase!!

Answer 1

$(3^{14} +7)^2 :3^{26}=3^{28}+7^2 :3^{26}=3^{28-26}+7^2=3^2+7^2=9+49=58$
$(11^{2011} * 2^{2010} ):[(22^2)^5]^{201}=11^{2011}*2^{2010}:22^{2010}= \\ =11^{2011}*(2:22)^{2010}=11^{2011}*(1:11)^{2010}=11^{2011}*1^{2010}:11^{2010}= \\ =11^{2011-2010}=11$