七年级数学有理数加减法简便计算练习题及参考答案C8

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+···+76。(2)(-1)+(-2)+(-3)+(-4)+···+(-560)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+397)+(+398)+(-399)+(-400)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1633)+(-1634)+(+1635)+(+1636)。(5)(-1)+(+2)+(-3)+(+4)+···+(-7333)+(+7334)。(6)(-2120)+(+2119)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+3293+(-3294)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,1863×1864)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,2199×2201)(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+···+76。解：原式=(1+76)+(2+75)+(3+74)+···+(38+39)=(1+76)*38=2926.(2)(-1)+(-2)+(-3)+(-4)+···+(-560)。解：原式=-(1+2+3+4+···+560)=-[(1+560)+(2+559)+(3+558)+···+(280+281)]=-[(1+560)*280]=-157080.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+397)+(+398)+(-399)+(-400)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+397)+(+398)+(-399)+(-400)]=(1+2-3-4)+(5+6-7-8)+···+(397+398-399-400)=-4+(-4)+···+(-4)=-4*100=-400.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1633)+(-1634)+(+1635)+(+1636)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-1633)+(-1634)+(+1635)+(+1636)]=(-1-2+3+4)+(-5-6+7+8)+···+(-1633-1634+1635+1636)=4+4+···+4=4*409=1636.(5)(-1)+(+2)+(-3)+(+4)+···+(-7333)+(+7334)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-7333)+(+7334)]=(-1+2)+(-3+4)···+(-7333+7334)=1+1+···+1=1*3667=3667.(6)(-2120)+(+2119)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-2120)+(+2119)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-2120+2119)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*1060=-1060.(7)1+(-2)+3+(-4)+···+3293+(-3294)。解：原式=[1+(-2)]+[3+(-4)]···+[3293+(-3294)]=(1-2)+(3-4)···+(3293-3294)=-1+(-1)+···+(-1)=-1*1647=-1647.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,1863×1864)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(1,1863)-eq\f(1,1864))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,1863)-eq\f(1,1864)=1-eq\f(1,1864)=eq\f(1863,1864).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,2199×2201)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(1,2199)-eq\f(1,2201))]=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(1,2199)-eq\f(1,2201))=eq\f(1,2)·(1-eq\f(1,2201))=eq\f(1100,2201).(10)eq\f(1,4)+eq\f(1,16)+eq\f(1,64)+eq\f(1,256)+eq\f(1,1024)。解：原式=H=eq\f(1,4)+eq\f(1,16)+eq\f(1,64)+eq\f(1,256)+eq\f(1,1024),两边同