几个数学难题,求助!1.已知a-b=4,ab+m^-6m+13=0,则ab+m的值为（）2.计算（1-1/2+1/3-1/4+...+1/2007-1/2008)/(1/1005+1/1006-1/1007+...+1/2008)=()3.已知：6/（n+1)(n+2)(n+3)(n+4)=a/(n+1)+b/(n+2)+c/(n+3)+d/(n+4),其中a,b

Question

几个数学难题,求助!
1.已知a-b=4,ab+m^-6m+13=0,则ab+m的值为（ ）
2.计算（1-1/2+1/3-1/4+...+1/2007-1/2008)/(1/1005+1/1006-1/1007+...+1/2008)=( )
3.已知：6/（n+1)(n+2)(n+3)(n+4)=a/(n+1)+b/(n+2)+c/(n+3)+d/(n+4),其中a,b,c,d是常数,则a+2b+3c+4d的值为（ ）

Answer 1

1.已知a-b=4,ab+m^2-6m+13=0,则ab+m的值为（ ）
(a-b)^2 = a^2+b^2-2ab =16
ab = (a^2+b^2)/2-8
代入有
(a^2+b^2)/2-8+m^2-6m+13=0
(a^2+b^2)/2+m^2-6m+5=0
(a^2+b^2-8)/2 + (m-3)^2 =0
( a^2+b^2 - (a-b)^2/2 )/2 + (m-3)^2 =0
( a^2+b^2 - (a-b)^2/2 )/2 + (m-3)^2 =0
( (a+b)^2/2 )/2 + (m-3)^2 = 0
很明显了吧,a+b=0,m=3
所以a=2,b=-2,m=3
ab+m = -1
2.好诡异的式子……1/1005+1/1006-1/1007+...+1/2008这里开头的加减符号错了吧?怎么连着1005,1006两个正的?
3.已知：6/（n+1)(n+2)(n+3)(n+4)=a/(n+1)+b/(n+2)+c/(n+3)+d/(n+4),其中a,b,c,d是常数,则a+2b+3c+4d的值为（ ）
a/(n+1)+b/(n+2)+c/(n+3)+d/(n+4)
=[a(n+2)(n+3)(n+4)+b(n+1)(n+3)(n+4)+c(n+1)(n+2)(n+4)+d(n+1)(n+2)(n+3)]/(n+1)(n+2)(n+3)(n+4)
显然看分子n^3的系数,知道a+b+c+d = 0
再看n^2的系数,例如a项为 (2+3+4)a = 9a,知道9a+8b+7c+6d =0 (1)
a+b+c+d = 0可以得到 10a+10b+10c+10d =0 (2)
(2)-(1) = a+2b+3c+4d = 0