Question

解方程:1/[x(x+1)]+1/(x+1)(x+2)+1/(x+2)(x+3)+....+1/(x+9)(x+10)=5/12

解方程:1/[x(x+1)]+1/(x+1)(x+2)+1/(x+2)(x+3)+....+1/(x+9)(x+10)=5/12

Answer 1

1/[x(x+1)]+1/(x+1)(x+2)+1/(x+2)(x+3)+....+1/(x+9)(x+10)=5/12
1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+9)-1/(x+10)=5/12;
1/x-1/(x+10)=5/12;
12(x+10)-12x=5x(x+10);
120=5x²+50x;
x²+10x-24=0;
(x+12)(x-2)=0;
x=2或x=-12；

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Answer 2

原式
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+1998)-1/(x+1999)
=1/x-1/(x+1999)
=1999/[x(x+1999)]

1/[x(x+1)]+1/(x+1)(x+2)+1/(x+2)(x+3)+....+1/(x+9)(x+10)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+9)-1/(x+10)
=1/x-1/(x+10)
=10/[x(x+10)] = 5/12
5 x(x+10) =120
x(x+10) =24
x² +10x-24 = 0
(x+12)(x-2) = 0
x=-12 或者 2

Answer 3

1/x(x＋1)＋1/(x＋1)(x＋2)＋1/(x＋2)(x＋3)＋……＋1/(x＋9)(x＋10)＝5/12
1/x－1/(x＋1)＋1/(x＋1)－1/(x＋2)＋1/(x＋2)－1/(x＋3)＋……＋1/(x＋9)－1/(x＋10)＝5/12
1/x－1/(x＋10)＝5/12
10/x(x＋10)＝5/12

5x²＋50x＝120
x²＋10x－24＝0
(x－2)(x＋12)＝0
x1＝2 x2＝-12

Answer 4

裂项啊 1/x-1/(x+10)=5/12 x=2

追问

能解说下不

追答

1/x-1/(1+x)=1/x(1+x)
楼下是对的，确实还有-12