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 calculati (a-b)^2021

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
=[27×2^2×5+(7×2)^2-7]÷(9)^3-1=
=(540+14^2-7)÷729-1=
=(540+196-7)÷729-1=
=729÷729-1=0

b=(3^17÷9^8)^2020÷(3^5)^404 =
=(3^17÷3^16)^2020÷(3^5)^404 =
=(3^1)^2020÷3^2020 =
=(3^2020÷3^2020 =1

(a-b)^2021 =(0-1)^2021=-1