Question

pls ajutor va roooggg​

Answer 1

(2^3)^35: (2^2*2^5)^15+[3^33-3^40: 3^20-(3^5)^2: 3^3^2]:3^2019^0=

2^105: (2^7)^15+(3^33-3^20-3^10:3^9):3=

2^105:2^105+(3^33-3^20-3):3=

2^0+3^32-3^19-1= 3^32-3^19