1a) 5x[70:35+2x(36:12+10]=
5x[2+2x(3+10)]=
5x[2+2x13]=
5x(2+26)=5x28=140
b) 729:3³+180:4+(42:3+12):2=
729:27+45+(14+12):2=
27+45+26:2=
72+13=85
2) y:4=c rest 7
folosim teorema impartirii cu rest
y=4c+7
daca c=1⇒y=4x1+7=11
daca c=2⇒y=4x2+7=15
daca c=3⇒y=4x3+7=19
daca c=4⇒y=4x4+7=23
daca x=5⇒y=4x5+7=27
nr.sunt:11,15,19,23,27
3)32=2⁵
128=2⁷
512=2x3²x7
4³⁵=(2²)³⁵=2⁷⁰
16⁷=(2⁴)⁷=2²⁸
1024=2⁴x3²x7