f(x) = x/(2x^2+7x-4)= (1/9)[1/(2x-1)]+(4/9)[1/(x+4)]= (-1/27){1/[1-(2/3)(x+1)]}+(4/27){1/[1+(1/3)(x+1)]}= (-1/27)Σ[(2/3)(x+1)]^n+(4/27)Σ[-(1/3)(x+1)]^n= …,|x+1|x/(2x^2+7x-4)=(1/9)[1/(2x-1)]+(4/9)[1/(x+4)]这是怎么过来的用待定系数法得到的。