Question

初三数学分式

(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

Answer 1

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

Answer 2

(1)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+2009)-1/(x+2011)=1/(x+1)-1/(x+2011)=2010/(x+1)(x+2011)
(2)(5x+4)/x(x+2)=A/x+B/(x+2)，A/x+B/(x+2)=[A(x+2)+Bx]/X(X+2),因此A+B=5,2A=4，随意A=2，B=3