七年级数学有理数加减法简便计算练习题及参考答案C7

七年级数学有理数加减法简便计算练习题及答案简要步骤姓名：班级：一、计算练习题(1)1+2+3+4+···+788。(2)(-1)+(-2)+(-3)+(-4)+···+(-636)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2265)+(+2266)+(-2267)+(-2268)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-2917)+(-2918)+(+2919)+(+2920)。(5)(-1)+(+2)+(-3)+(+4)+···+(-7059)+(+7060)。(6)(-5852)+(+5851)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+3461+(-3462)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,2004×2005)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,101×103)(10)eq\f(1,4)+eq\f(1,16)+eq\f(1,64)+eq\f(1,256)+eq\f(1,1024)。二、详细步骤(1)1+2+3+4+···+788。解：原式=(1+788)+(2+787)+(3+786)+···+(394+395)=(1+788)*394=310866.(2)(-1)+(-2)+(-3)+(-4)+···+(-636)。解：原式=-(1+2+3+4+···+636)=-[(1+636)+(2+635)+(3+634)+···+(318+319)]=-[(1+636)*318]=-202566.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2265)+(+2266)+(-2267)+(-2268)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+2265)+(+2266)+(-2267)+(-2268)]=(1+2-3-4)+(5+6-7-8)+···+(2265+2266-2267-2268)=-4+(-4)+···+(-4)=-4*567=-2268.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-2917)+(-2918)+(+2919)+(+2920)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-2917)+(-2918)+(+2919)+(+2920)]=(-1-2+3+4)+(-5-6+7+8)+···+(-2917-2918+2919+2920)=4+4+···+4=4*730=2920.(5)(-1)+(+2)+(-3)+(+4)+···+(-7059)+(+7060)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-7059)+(+7060)]=(-1+2)+(-3+4)···+(-7059+7060)=1+1+···+1=1*3530=3530.(6)(-5852)+(+5851)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-5852)+(+5851)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-5852+5851)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*2926=-2926.(7)1+(-2)+3+(-4)+···+3461+(-3462)。解：原式=[1+(-2)]+[3+(-4)]···+[3461+(-3462)]=(1-2)+(3-4)···+(3461-3462)=-1+(-1)+···+(-1)=-1*1731=-1731.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,2004×2005)解：原式=1-eq\f(1,2)+(eq\f(1,2)-eq\f(1,3))+(eq\f(1,3)-eq\f(1,4))+···+(eq\f(1,2004)-eq\f(1,2005))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,2004)-eq\f(1,2005)=1-eq\f(1,2005)=eq\f(2004,2005).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,101×103)解：原式=eq\f(1,2)·[(1-eq\f(1,3))+(eq\f(1,3)-eq\f(1,5))+(eq\f(1,5)-eq\f(1,7))+···+(eq\f(1,101)-eq\f(1,103))]=eq\f(1,2)·(1-eq\f(1,3)+eq\f(1,3)-eq\f(1,5)+eq\f(1,5)-eq\f(1,7)+···+eq\f(1,101)-eq\f(1,103))=eq\f(1,2)·(1-eq\f(1,103))=eq\f(51,103).(10)eq\f(1,4)+eq\f(1,16)+eq\f(1,64)+eq\f(1,256)+eq\f(1,1024)。解：原式=Q=eq\f(1,4)+eq\f(1,16)+eq\f(1,64)+eq\f(1,256)+eq\f(1,1024),两边