Question

rezolvarea va rog la b si c

Answer 1

b)1.2(3)-{[0.(6)-1_6]+(0.15-1_4)}=
=1.2(3)-{[6_9-1_6]+0.15-1_4)]=
=1.2(3)-{[12_18-3_18]+0.15-1_4)]=
=1.2(3)-{[9_18]+0.15-1_4)]=
=1.2(3)-[(9_18)+0.15-1_4]=
=1.2(3)-(9_18+0.15-1_4)=
=1.2(3)-(9_18+15_100-1_4)=
=1.2(3)-(90_100+15-100-25_100)=
=1.2(3)-(80_100)=
=123_109-80_100=

Answer 2

b)1,2(3)-{[0,(6)-1/6]+(0,15-1/4)}=
=(123-12)/90-[(6/9-1/6)+(15/100-1/4)]=
=112/90-[(2/3-1/6)+(3/20-1/4)]=
=61/45-[(4/6-1/6)+(3/20-5/20)]=
=61/45-(3/6-2/20)=
=61/45-(1/2-1/10)=
=61/45-(5/10-1/10)=
=61/45-4/10=
=61/45-2/5=
=61/45-18/45=
=43/45


c)[1/2(24)-2 int5/33]-(15/33-11-15)=
=[(1224-12)/990-(2•33+5)/33]-[(15•3)/165-(11•11)/165=
=(1212/990-71/33)-(45/165-121/165)=
=(404/330-71/33)-(-76/165)=
=(404/330-710/330)+76/165=
=306/330+76/165=153/165+76/165=
=229/165=
=1 intreg 64/165