Question

2^100-[2^50*2^48+(2^10*2^15)^5:2^27+(5^76:5&%-3)^9*2^8+2^98]= 2^100-[2^98+(2^25)^5:2^27+(5-3)^90*2^8+2^98=2^100-(2^98+2^125:2^27+2^90*2^8+2^98)=2^100-(2^98+2^98+2^98+2^98)= va rog frumos daca stie cineva cum trebuie facut in continuare

Answer 1

2¹⁰⁰-[2⁹⁸+(2²⁵)⁵:2²⁷+(5-3)⁹⁰*2⁸+2⁹⁸] =2¹⁰⁰- (2⁹⁸+ 2¹²⁵: 2²⁷+2⁹⁰*2⁸ +2⁹⁸)=

=2¹⁰⁰-(2⁹⁸ +2⁹⁸+2⁹⁸+2⁹⁸)= 2¹⁰⁰ - 2⁹⁸*(1+1+1+1)= 2¹⁰⁰ - 2⁹⁸*4= 2¹⁰⁰ - 2⁹⁸*2²=

=2¹⁰⁰ -2¹⁰⁰ = 2¹⁰⁰ *( 1-1)= 2¹⁰⁰*0=0