(1) 化简：2/(x+1)(x+3)+2/(x+3)(x+5)+2/(x+5)(x+7)+...+2/(x+2009)(x+20_数学解答

(1) 化简：2/(x+1)(x+3)+2/(x+3)(x+5)+2/(x+5)(x+7)+...+2/(x+2009)(x+2011)
(2) 若对于一切不为0且不为-2的实数x,(5x+4)/x(x+2)=A/x+B(x+2),求常数A,B的值

人气：128 ℃　时间：2020-02-05 17:52:09

解答

2/(x+1)(x+3)+2/(x+3)(x+5)+2/(x+5)(x+7)+...+2/(x+2009)(x+2011)
=1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+1/(x+5)-1/(x+7)+...+1/(x+2009)-1/(x+2011)
=1/(x+1)-1/(x+2011)
=(x+2011)/(x+1)(x+2011)-(x+1)/(x+1)(x+2011)
=(x+2011-x-1)/(x+1)(x+2011)
=2010/(x+1)(x+2011)
(5x+4)/x(x+2)=A/x+B/(x+2)
(5x+4)/x(x+2)=A(x+2)/x(x+2)+Bx/x(x+2)
(5x+4)/x(x+2)=[A(x+2)+Bx]/x(x+2)
(5x+4)/x(x+2)=[Ax+2A+Bx]/x(x+2)
(5x+4)/x(x+2)=[(A+B)x+2A]/x(x+2)
A+B=5
2A=4
解得
A=2
B=3