Question

Punctul (a ) si (d)
Ajutor

Answer 1

a) [3,8(3)-2,(54)×11/28]:2,(3)=
3,8(3)=(383-38)/90=345/90=23/6
2,(54)=(254-2)/99=252/99=28/11
2,(3)=(23-2)/9=21/9=7/3
=[23/6-28/11×11/28]:7/3=
=(23/6-1):7/3=
=(23-6)/6 ×3/7=
=17/6×3/7=17/14

d)[0,278:13,9+(2-0,47):3/20]:102,2+3,4×1 4/17=
=[278/1000 :139/10+153/100 ×20/3]:1022/10+34/10×21/17=
=(278/1000 ×10/139+51/5) ×10/1022+42/10=
=(2/100+51/5) ×10/1022+42/10=
=(2+20×51)/100 ×10/1022+42/10=
=1022/100 ×10/1022+42/10=
=1/10+42/10=43/10