七年级数学有理数加减法简便计算练习题及参考答案C2

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+···+168。(2)(-1)+(-2)+(-3)+(-4)+···+(-172)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2513)+(+2514)+(-2515)+(-2516)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1877)+(-1878)+(+1879)+(+1880)。(5)(-1)+(+2)+(-3)+(+4)+···+(-4945)+(+4946)。(6)(-8450)+(+8449)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+609+(-610)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,3852×3853)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,4401×4403)(10)eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776)。二、详细步骤(1)1+2+3+4+···+168。解：原式=(1+168)+(2+167)+(3+166)+···+(84+85)=(1+168)*84=14196.(2)(-1)+(-2)+(-3)+(-4)+···+(-172)。解：原式=-(1+2+3+4+···+172)=-[(1+172)+(2+171)+(3+170)+···+(86+87)]=-[(1+172)*86]=-14878.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2513)+(+2514)+(-2515)+(-2516)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+2513)+(+2514)+(-2515)+(-2516)]=(1+2-3-4)+(5+6-7-8)+···+(2513+2514-2515-2516)=-4+(-4)+···+(-4)=-4*629=-2516.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1877)+(-1878)+(+1879)+(+1880)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-1877)+(-1878)+(+1879)+(+1880)]=(-1-2+3+4)+(-5-6+7+8)+···+(-1877-1878+1879+1880)=4+4+···+4=4*470=1880.(5)(-1)+(+2)+(-3)+(+4)+···+(-4945)+(+4946)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-4945)+(+4946)]=(-1+2)+(-3+4)···+(-4945+4946)=1+1+···+1=1*2473=2473.(6)(-8450)+(+8449)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-8450)+(+8449)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-8450+8449)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*4225=-4225.(7)1+(-2)+3+(-4)+···+609+(-610)。解：原式=[1+(-2)]+[3+(-4)]···+[609+(-610)]=(1-2)+(3-4)···+(609-610)=-1+(-1)+···+(-1)=-1*305=-305.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,3852×3853)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(1,3852)-eq\f(1,3853))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,3852)-eq\f(1,3853)=1-eq\f(1,3853)=eq\f(3852,3853).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,4401×4403)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(1,4401)-eq\f(1,4403))]=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(1,4401)-eq\f(1,4403))=eq\f(1,2)·(1-eq\f(1,4403))=eq\f(2201,4403).(10)eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776)。解：原式=L=eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776),两边同时乘