數學Partial Functions 求詳解?

Sol
(d)
設 (x^2+4x+8)/(x^3+9x^2+27x+27)=a/(x+3)+b/(x+3)^2+c/(x+3)^3
x^2+4x+8=a(x+3)^2+b(x+3)+c
Set x=－3
9－12+8=c
c=5
x^2+4x+3=a(x+3)^2+b(x+3)
x+1=a(x+3)+b
Set x=-3
－3+1=b
b=－2
a=1
So
(x^2+4x+8)/(x^3+9x^2+27x+27)=1/(x+3)－2/(x+3)^2+5/(x+3)^3
(g)
設 (x^3+9x^2+9x+15)/[(x^2+1)(x^2+4x+4)]=(ax+b)/(x+1)^2+c/(x+2)+d/(x+2)^2
x^3+9x^2+9x+15=(ax+b)*(x+2)^2+c(x^2+1)(x+2)+d(x^2+1)
Set x=-2
－8+36－18+15=d(2^2+1)
5d=25
d=5
x^3+9x^2+9x+15=(ax+b)*(x+2)^2+c(x^2+1)(x+2)+5(x^2+1)
x^3+4x^2+9x+10=(ax+b)*(x+2)^2+c(x^2+1)(x+2)
x^2+2x+5=(ax+b)(x+2)+c(x^2+1)
Set x=－2
4－4+5=c*5
c=1
x^2+2x+5=(ax+b)(x+2)+(x^2+1)
2x+4=(ax+b)(x+2)
a=0，b=2
So
(x^3+9x^2+9x+15)/[(x^2+1)(x^2+4x+4)]=2/(x+1)^2+1/(x+2)+5/(x+2)^2

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