Question

calculati, respectand regulile invatate.
a.) [26+117×9-(832÷2-531÷9)]×10
b) 30+4[36÷9+3×(20+8×42÷2)]
c) 2100÷(225-500÷4)+(84÷2×0)]×100÷10÷10
d) 9×(101×1+111÷1)-[85-7×(36÷9+63÷9)]
e) 3+6×[30+6×(3+5×6)-28]-3
f) 2+2×[2+3×(2+202÷2-306÷3)]

Answer 1

a) 1. 832/2=416 , 531/9=59 , 416-59=357
2. 117x9=1053 , 1053+26=1079
3. 1079-357=722 
4. 722x10=7220

Regula :
1) prima operatie se efectuiaza cea din paranteza , in cazul dat prima operatie din se va fi efectuata impartirea apoi scaderea .
2) oa doua operatie din paranteza patrata , in cazul dat in primul rind se efectuiaza inmultirea apoi adunarea.
3)apoi se efectuiaza oeratia din inafara parantezelor 

Answer 2

îia)[26+117x9-(832:2-531:9)]x10=
[26+1053-(416-59)]x10=
[26+1053-357]x10=
[1079-357]x10=
722x10=7220

b)30+4x[36:9+3x(20+8x42:2)]=
30+4x[4+3x(20+336:2)]=
30+4x[4+3x(20+168)]=
30+4x[4+3x188]=
30+4x[4+564]=
30+4x568=
30+2272=2302

c)2100:(225-500:4)+(84:2x0)]x100:10:10=
2100:[(225-125)+(42x0)]x100:10:10=
2100:[100+0]x100:10:10=
2100:100x100:10:10=
21x100:10:10=
2100:10:10=
210:10=21

d)9x(101x1+111:1)-[85-7x(36:9+63:9)]=
9x(101+111)-[85-7x(4+7)]=
9x212-[85-7x11]=
1908-[85-77]=
1908-8=1900

e)3+6x[30+6x(3+5x6)-28]-3=
3+6x[30+6x(3+30)-28]-3=
3+6x[30+6x33-28]-3=
3+6x[30+198-28]-3=
3+6x[228-20]-3=
3+6x208-3=
3+1248-3=
1251-3=1248

f)2+2x[2+3x(2+202:2-306:3)]=
2+2x[2+3x(2+101-102)]=
2+2x[2+3x(103-102)]=
2+2x[2+3x1]=
2+2x[2+3]=
2+2x5=
2+10=12

Mult succes la școală!