Question

Fie a=[27×2^2×5+(7×2)^2-7]:(9)^3^2^0-[1^3^2^5+(2^2)^3^0^7]^0 si b=(3^17:9^8)^2020:(3^5)^404

Answer 1

a=[27×2^2×5+(7×2)^2-7]:(9)^3^2^0-[1^3^2^5+(2^2)^3^0^7]^0

a = (27×4×5+14^2 - 7) : 9^3 - 1

paranteza patrata fiind la puterea 0, nu se mai calculeaza, stiind ca orice numar ridicat la putere 0 este 1

a = ( 108 × 5 + 196 - 7) : 729 - 1

a = ( 540 + 196 - 7) : 729 - 1

a = ( 736 - 7) :729 - 1

a = 729 : 729 - 1

a = 1 - 1

a = 0
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b = (3^17:9^8)^2020:(3^5)^404

b = [ 3^17 : (3^2)^8]^2020 : 3^2020

b = ( 3^17 : 3^16)^2020 : 3^2020

b = 3^2020 : 3^2020

b = 1
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