Question

Calculați : 1) 3^28:9^12+3^3^1-3^0•[729:3^5+3^3:(3•3^2-18^1)]=
2) 2^30:2^25+2^4^0-2^1•[128:2^5+2^3:(2•2^4-24^1)]=
3) 7^30:49^15+7^2^0+7^1•[2401:7^3+7^2-336^1)]= VĂ ROG FRUMOS AJUTA-ȚIMĂ !!!

Answer 1

1) 3^28:9^12+3^3^1-3^0•[729:3^5+3^3:(3•3^2-18^1)]=
= 3^28:3^24+3^3-[3^6:3^5+3^3:(3^3-3^2•2)]=
= 3^4+3^3-{3+3^3:[3^2(3-2)]=
= 3^4+3^3-(3+3^3:3^2)=
= 3^4+3^3-(3+3)=
= 3^4+3^3-2•3=
=3(3^3+3^2-2)=
=3(27+9-2)=
=68

2) 2^30:2^25+2^4^0-2^1•[128:2^5+2^3:(2•2^4-24^1)]=
=2^5+2^1-2•[2^7:2^5+2^3:(2^5-2^3•3)]=
=2^5+2^1-2•{2^2+2^3:[2^3(2^2-3)]}=
=2^5+2^1-2•(2^2+2^3:2^3)=
=2^5+2^1-2•(2^2+1)=
=2^5+2^1-2•5=
=2•(2^4+2^0-10)=
=2•(16+1-10)=
=2•7=14

 3) 7^30:49^15+7^2^0+7^1•[2401:7^3+7^2-336^1)]=
= 7^30:7^30+7^1+7•[7^4:7^3+7^2-336)]=
= 1+7+7•(7+49-336)=
=8-7•280=
=8-1960=
=1952