给我出20道有理数加减法,给高分,要步骤和答案

问题描述:

给我出20道有理数加减法,给高分,要步骤和答案
跪求呀,10道也可以!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!混和计算也可以!!!拜托了!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

①-7/5 +(+ 1/10) =-1.4+0.1=-1.3       ②90-(-3)=90+3=93③-0.5-(-3 )+2.75-(+7 ) =0.5+3+2.75-7=6.25-7=-0.75 ④ (-2.8)+(+1.9)=-2.8+1.9=-0.9就类似的,换换娄 -09+1.1-(0.2)+1...还有么?(-38)+52+118+(-62)= (-38-62)+(52+118)=-100+170=70(-32)+68+(-29)+(-68)= (-32-29)+(68-68)=-61(-21)+251+21+(-151)= (-21+21)+(251-251)=012+35+(-23)+0= 37-23=14(-6)+8+(-4)+12 = 2-4+12=-2+12=1027+(-26)+33+(-27) =1+33-27=34-27=739+[-23]+0+[-16]= 16+0-16=16-16=0[-18]+29+[-52]+60= 18+23+52+60=41+52+60=93+60=153[-3]+[-2]+[-1]+0+1+2=3+2+1+0+1+2=9(-301)+125+301+(-75)= (-301+301)+(125-75)=0+50=501-4/9 = 5/91-7/10=3/10 8/15-5= -4又7/157-15=-8 2/8-5/8=-3/8 11/12-10/12= -1/12 42-(3+3 )= 42-6=361/3- 7/12-7/18=12/36-21/36-14/36=(12-21-14)/36=-23/36 1 -1/3-1 1/5 =2/3-1/5=10/15-3/15=7/15 10-7/10= 9又3/105/24+3/8 =5/24+9/24=14/24=7/8 1-3/5=2/5 4.39x1/13x2/3 =3x2/3=21+(-2)+(-3)+4+5+(-6)+(-7)+8+9+(-10)+(-11)+12 =-1-3+4+5-6-7+8+9-10-11+12=-4+4+5-6-7+8+9-10-11+12=5-6-7+8+9-10-11+12=-1-7+8+9-10-11+12=-8+8+9-10-11+12=9-10-11+12=-1-11+12=-12+12=0(-7分之1)+(-7分之2)+1又7分之3 =-7分之3+1又7分之3=1