f(x)/(x+2)=g(x)+1 (1)
f(x)/(x+3)=g(x)-1 (2)
(1)-(2),
f(x)/(x+2)-f(x)/(x+3)=2
[(x+3-x-2)f(x)]/[(x+2)(x+3)]=2
f(x)/[(x+2)(x+3)]=2
所以余式为0