Question

已知分式方程(2X-3)／(X²+X)=A／(X+1)+(B)／(X),已知A、B为常数,求A-B的值

Answer 1

(2x-3)/(x²+x)=A/(x+1)+B/x

(2x-3)/[x(x+1)]=[Ax+B(x+1)]/[x(x+1)]

(2x-3)/[x(x+1)]=[(A+B)x+B]/[x(x+1)]

A+B=2

B=-3

解得A=5 B=-3

A-B=5-(-3)=8