Question

解方程：1/（x+1)(x+2)+1/(x+3)(x+4)+……+1/(x+2010)(x+2011)=(2x+4019)/(3x+6033)

Answer 1

∵1/（x+1)(x+2)=1/（x+1)-1/(x+2)
∴1/（x+1)(x+2)+1/(x+3)(x+4)+……+1/(x+2010)(x+2011)
=1/（x+1)-1/(x+2)+1/(x+3)-1/(x+4)+.......+1/(x+2010)-1/(x+2011)
=1/(x+1)-1/(x+2011)=(2x+4019)/(3x+6033)
∴1/(x+1)=(2x+4019)/(3x+6033)+3/(3x+6033)
∴1/(x+1)=2x+4022/(3x+6033)
即1/(x+1)=2/3
∴x=1/2

Answer 2

注意到：1/[n(n+1)]=1/n-1/(n+1)
x≠-1,-2....-2011
左边=1/(x+1)-1/(x+2011)
=2010/[(x+1)(x+2011)=(2x+4019)/3*1/(x+2011)=右边
2x^2+4021x-2011=0
(2x-1)(x+2011)=0
x=1/2或x=-2011(舍）
所以：
x=1/2

Answer 3

1/（x+1)- 1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)- 1/(x+4)+……+1/(x+2010)- 1/(x+2011)=(2x+4019)/(3x+6033)
1/（x+1)- 1/(x+2011)=(2x+4019)/(3x+6033)
1/（x+1) =(2x+4019)/(3x+6033)+ 1/(x+2011)
1/（x+1) =(2x+4019+3)/[3(x+2011)]
1/（x+1) =[2(x+2011)] /[3(x+2011)]
1/（x+1) =2/3,
X=1/2.

Answer 4

你的题目有问题，第一个分子式是1、2，是奇数、偶数交替出现的，你等号前面的怎么变成偶数、奇数了呢

Answer 5

结果是1/2 以下是matlab程序代码

function ssans=syyyy()
sans='0';
for si=1:2010
sa=['(x+',num2str(si),')'];
san=['(x+',num2str(si+1),')'];
sa=['1/(',sa,'*',san,')'];
sans=[sans,'+',sa];
end
syms x;
sans=eval(sans);
sans=simplify(sans,'100');
sans=char(sans);
sans=[sans,'-(2*x+4019)/(3*x+6033)'];
ssans=solve(sans);
end