求一个裂项公式的通式求下面这2个裂项公式的的结果,恳请大人指点1/(ax+b)(cx+d)=?mx/(ax+b)(cx+d)=?a,b,c,d,m均为常数

Question

求一个裂项公式的通式
求下面这2个裂项公式的的结果,恳请大人指点
1/(ax+b)(cx+d)=?
mx/(ax+b)(cx+d)=?
a,b,c,d,m均为常数

Answer 1

设1/(ax+b)(cx+d)=k/(ax+b)+n/(cx+d)=[k(cx+d)+n(ax+b)]/(ax+b)(cx+d)
=[(kc+na)x+(kd+nb)]/(ax+b)(cx+d)
所以kc+na=0,kd+nb=1
所以n=-c/(ad-bc)
k=a/(ad-bc)
所以1/(ax+b)(cx+d)=[a/(ad-bc)]/(ax+b)-[c/(ad-bc)]/(cx+d)
设mx/(ax+b)(cx+d)=k/(ax+b)+n/(cx+d)=[k(cx+d)+n(ax+b)]/(ax+b)(cx+d)
=[(kc+na)x+(kd+nb)]/(ax+b)(cx+d)
所以kc+na=m,kd+nb=0
所以n=-dm/(bc-ad)
k=bm/(bc-ad)
所以mx/(ax+b)(cx+d)=[bm/(bc-ad)]/(ax+b)-[dm/(bc-ad)]/(cx+d)