Question

20x{17^2:289+5x[(3^11x5)^10:(3^110x5^9)+(2^2-3)^2020]}dau coroana​

Answer 1

20[17^2:17^2+5* (3^11*5)^10/3^110*5^9 +(4-3)^2020]
20(1+ 3^110*5^10/3^110*5^8 +1^2020)
20(1+5^2+1)
20(1+25+1)
20*27=540