Question

求高人指教

Answer 1

f(x)=(x+a)(x+b)(x-2)-4=(x-3)g(x)+20
所以就有：(3+a)(3+b)=24
3+a>=4，3+b>3+a>=4
因此,3+a=4，3+b=6
故a=1，b=3
\frac{3x+9}{f(x)+4}+\frac{4}{x^2+6x+5}
=\frac{3(x+3)}{(x+1)(x+3)(x-2)}+\frac{4}{(x+1)(x+5)}
=\frac{3}{(x+1)(x-2)}+\frac{4}{(x+1)(x+5)}
=\frac{1}{x+1}\left(\frac{3}{x-2}+\frac{4}{x+5}\right)
=\frac{1}{x+1}\frac{3(x+5)+4(x-2)}{(x-2)(x+5)}
=\frac{1}{x+1}\frac{7(x+1)}{(x-2)(x+5)}
=\frac{7}{(x-2)(x+5)}