若在关于x的恒等式(Mx+N)/(x2+x-2)=2/(x+a)-c/(x-b)中,(Mx+N)/(x2+x-2)为最简分式,且有a＞b,a+b=c则N=.

Question

若在关于 x 的恒等式 (Mx+N)/(x2+x-2) = 2/(x+a) - c/(x-b)中,(Mx+N)/(x2+x-2) 为最简分式,
且有a＞b,a+b=c 则N=_______.

Answer 1

∵ x²+x-2=(x+2)(x-1)
∵ a>b
∴ a=2,b=-1
(mx+n)/(x²+x-2)=[2/(x+2)]-[c/(x-1)]
∴(mx+n)/(x²+x-2)=[2(x-1)-c(x+2)]/(x²+x-2)
即 (mx+n)/(x²+x-2)=[(2-c)x+(-2-2c)]/(x²+x-2)
∴ m=2-c ①
n=-2-2c ②
c=a+b=1 ③
∴ m=1,n=-4