a) 3^101:9^50=3^101:3^100
=3^1=3
(3^57:9^25×3^10):3^15=(3^57:3^100×3^10):3^15=
=(3^57:3^110):3^15=
=3^53:3^15=
=3^37
b) 3^(7+10-15)=3^2=9
c) 2×4×2^43-2^46+[9^74:3^43-(3^5)^6]^0=2×2^2×2^43-2^46+1=2^46×(1-1)+1=
=2^46×0+1=1
[9^74:3^43-(3^15)^6]^0 (orice nr. la puterea 0 este egal cu 1).
Succes!
Coroană pls?
