Question

求解，详细过程，谢谢。不太懂这种类型的题目怎么做ʕ•ﻌ•ʔ

Answer 1

y=1/(x^2+3x-4)
=1/[(x-1)(x+4)]
=1/5[1/(x-1)-1/(x+4)]
y'=1/5[-1/(x-1)^2+1/(x+4)^2]
=1/5[(-1)/(x-1)^2+(-1)^(1+1)/(x+4)^2]
y"=1/5[(-1)(-2)/(x-1)^3+(-1)^(1+1)(-2)/(x+4)^2]
=1×2/5[(-1)^2/(x+3)^3+(-1)^(2+1)/(x+4)^3]
y"'=1×2/5[(-1)^2(-3)/(x+3)^4+(-1)^(2+1)(-3)/(x+4)^4]
=1×2×3/5[(-1)^3/(x+3)^4+(-1)^(3+1)/(x+4)^4]
...
y^(n)=1×2×3×...×n/5[(-1)^n/(x+3)^(n+1)+(-1)^(n+1)/(x+4)^(n+1)]
=n!/5[(-1)^n/(x+3)^(n+1)+(-1)^(n+1)/(x+4)^(n+1)]
注：^——表示次方。

追问

求解

1/5是怎么算出来的

追答

∵1/(x-1)-1/(x+4)
=[(x+4)-(x-1)]/[(x-1)(x+4)]
=5/[(x-1)(x+4)]
∴1/[(x-1)(x+4)]
=1/5[1/(x-1)-1/(x+4)]

追问

谢谢