七年级数学有理数加减法简便计算练习题及参考答案C1

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+···+1332。(2)(-1)+(-2)+(-3)+(-4)+···+(-424)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+1169)+(+1170)+(-1171)+(-1172)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-3005)+(-3006)+(+3007)+(+3008)。(5)(-1)+(+2)+(-3)+(+4)+···+(-4345)+(+4346)。(6)(-4824)+(+4823)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+8051+(-8052)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,3160×3161)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,587×589)(10)eq\f(1,2)+eq\f(1,4)+eq\f(1,8)+eq\f(1,16)+eq\f(1,32)。二、详细步骤(1)1+2+3+4+···+1332。解：原式=(1+1332)+(2+1331)+(3+1330)+···+(666+667)=(1+1332)*666=887778.(2)(-1)+(-2)+(-3)+(-4)+···+(-424)。解：原式=-(1+2+3+4+···+424)=-[(1+424)+(2+423)+(3+422)+···+(212+213)]=-[(1+424)*212]=-90100.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+1169)+(+1170)+(-1171)+(-1172)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+1169)+(+1170)+(-1171)+(-1172)]=(1+2-3-4)+(5+6-7-8)+···+(1169+1170-1171-1172)=-4+(-4)+···+(-4)=-4*293=-1172.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-3005)+(-3006)+(+3007)+(+3008)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-3005)+(-3006)+(+3007)+(+3008)]=(-1-2+3+4)+(-5-6+7+8)+···+(-3005-3006+3007+3008)=4+4+···+4=4*752=3008.(5)(-1)+(+2)+(-3)+(+4)+···+(-4345)+(+4346)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-4345)+(+4346)]=(-1+2)+(-3+4)···+(-4345+4346)=1+1+···+1=1*2173=2173.(6)(-4824)+(+4823)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-4824)+(+4823)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-4824+4823)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*2412=-2412.(7)1+(-2)+3+(-4)+···+8051+(-8052)。解：原式=[1+(-2)]+[3+(-4)]···+[8051+(-8052)]=(1-2)+(3-4)···+(8051-8052)=-1+(-1)+···+(-1)=-1*4026=-4026.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,3160×3161)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(1,3160)-eq\f(1,3161))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,3160)-eq\f(1,3161)=1-eq\f(1,3161)=eq\f(3160,3161).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,587×589)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(1,587)-eq\f(1,589))]=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(1,587)-eq\f(1,589))=eq\f(1,2)·(1-eq\f(1,589))=eq\f(294,589).(10)eq\f(1,2)+eq\f(1,4)+eq\f(1,8)+eq\f(1,16)+eq\f(1,32)。解：原式=H=eq\f(1,2)+eq\f(1,4)+eq\f(1,8)+eq\f(1,16)+eq\