Question

{2×3^2-3^2+3×[3+3×(3^3-3^5÷3^3)]}÷(2017^0+0^2017+2^1)​

Answer 1

{2×3^2-3^2+3×[3+3×(3^3-3^5÷3^3)]}÷(2017^0+0^2017+2^1) =

= {2*9-9+3*[3+3*(3^3 - 3^2)]} : (1+0+2) =

= {18-9+3*[3+3*3^2*(3-1)]} : 3 =

= [9+3*(3+3^3*2)] : 3 =

= [9+3*(3+27*2)] : 3 =

= (9+3*57) : 3 =

= (9 + 171) : 3 =

= 180:3 =

= 60