a) 1^2020-2020^0+3^2-2^3=
1-1+9-8=1
b)2^12:2^10+5^2-3^2+2018^0=
2^2+25-9+1=
29-9+1=
21
c)2019^0+1^2019+2018^1+0^2018-2017^1=
1+1+2018+0-2017=
2020-2017=
3
d)[(2^3+2^4+2^5+2^6):(2^2+2^3)]*(1+3+3^2+3^3)=
[(8+16+32+64):(4+8)]*(1+3+9+27)=
(120:12)*40=
10*40=
400
la d) n-am folosit factorul comun, dar daca iti trebuie rezolvarea cu el......
