Question

如果{x+3}{x+4}分之2x+1=x-3分之A+x+4分之B,则常数A={ },B={ }

Answer 1

(2x+1)/(x-3)(x+4)=[A/(x-3)]+[B/(x+4)]

因为 [A/(x-3)]+[B/(x+4)]=A(x+4)/(x-3)(x+4)+B(x-3)/(x-3)(x+4)

=[A(x+4))+B(x-3)]/(x-3)(x+4)

=[（A+B）x+4A-3B]/(x-3)(x+4)

所以 (2x+1)/(x-3)(x+4)=[（A+B）x+4A-3B]/(x-3)(x+4)

即 (2x+1）=[（A+B）x+4A-3B]

所以 A+B=2且4A-3B=1

解这个方程组得

A=1,B=1