Question

ajutor va rog frumos rapid

Answer 1

a=2^59-(2^2)^29-(2^3)^19 a=2^59-2^58-2^57 a=2^57(2^2-2^1-1) a=2^57*1 a=2^57 b=3^39-2* 3^2*(3^3)^12 b=3^39-2*3^2*3^36 b=3^39-2*3^38 b=3^38(3-2*1) b=3^38*1 b=3^38 a=(2^19)^3 si b=(3^19)^2 =>a>b b) a=3^59-(3^2)^29-5*(3^3)^19 a=3^59-3^58-5*3^57 a=3^57(3^2-3-5*1) a=3^57*1 a=3^57 b=5^39-4*(5^2)^19 b=5^39-4*5^38 b=5^38(5-4*1) b=5^38*1 b=5^38 a=(3^19)^3 si b=(5^19)^2; a