Question

procedimiento plis xfa
de todos o los q puedan

Answer 1

ㅡ21ㅡ(ㅡ7)+4.(ㅡ7)ㅡ12 =

ㅡ21+7ㅡ28ㅡ12=

ㅡ14ㅡ40= ㅡ54

12ㅡ36+15.ㅡ3ㅡ6.(ㅡ9)=

12ㅡ36ㅡ45+54= ㅡ15

ㅡ5+7ㅡ(3+2.(ㅡ3))=

ㅡ5+7ㅡ(3ㅡ6)=

ㅡ5+7ㅡ(ㅡ3)=

ㅡ5+7+3= 5

[ㅡ23+(2.(ㅡ8)ㅡ5)+12]ㅡ5=

[ㅡ23+(ㅡ16ㅡ5)+12]ㅡ5=

[ㅡ23ㅡ21+12]ㅡ5=

[ㅡ44+12]ㅡ5=

ㅡ32ㅡ5= ㅡ37

[12.(ㅡ3)+8ㅡ12ㅡ(ㅡ6+2)]+2=

[ㅡ36+8ㅡ12ㅡ(ㅡ4)]+2=

[ㅡ36+8ㅡ12+4]+2=

[ㅡ48+12]+2=

ㅡ36+2= ㅡ34