Question

C)3^40:[3^20×3^18+(3^5×3^14)^2+(4^20:4^19-1)^38]= d)8^33:[4^32×2^34+(2^5×2^20)^5:(16×2^23)+(75:75-2^0)×32×4]= va rog f

Answer 1

c)3^40:[3^20×3^18+(3^5×3^14)^2+(4^20:4^19-1)^38]=
=3^40:[3^38+(3^19)^2+(4^1-1)^38]=
=3^40:(3^38+3^38+3^38)=
=3^40:[3^38(1+1+1)]=
=3^40:(3^38×3)=
=3^40:3^39=3

 d)8^33:[4^32×2^34+(2^5×2^20)^5:(16×2^23)+(75:75-2^0)×32×4]=
=2^99:[2^64×2^34+(2^25)^5:(2^4×2^23)+(1-1)×32×4]=
=2^99:[2^98+2^125:2^27+0]=
=2^99:(2^98+2^98)=
=2^99:[2^98(1+1)]=
=2^99:(2^98×2)=
=2^99:2^99=1