求y=(x^3-1)/(x^2-2x-3)的n阶导数其实化成1/(ax+b)形式我也明白,但是问题是,如何化到这一步?只要给我讲一讲(x^3-1)/(x^2-2x-3)是如何化成x+2+(7x+5)/(x+1)(x+3)就可以了,后面的我自己会.
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求y=(x^3-1)/(x^2-2x-3)的n阶导数
其实化成1/(ax+b)形式我也明白,但是问题是,如何化到这一步?
只要给我讲一讲(x^3-1)/(x^2-2x-3)是如何化成x+2+(7x+5)/(x+1)(x+3)就可以了,后面的我自己会.
分子3次,分母2次,且系数比是1,所以
原式=x+(Ax^2+Bx+C)/(x^2-2x-3)=(x^3 + (A-2)x^2 +(B-3)x +C)/(x^2-2x-3) 所以A=2,B=3,C=-1
所以原式=x+(Ax^2+Bx+C)/(x^2-2x-3)=x+(2x^2+3x-1)/(x^2-2x-3)
同理有原式=x+2+(7x+5)/(x^2-2x-3)=x+2+(7x+5)/(x+1)(x-3)
其实就是初中时恒等变形的基本功,另外,你分母分解因式的时候写错了,应该是(x+1)(x-3)
原式=(x^3-2x^2-3x+2x^2+3x-1)/(x^2-2x-3)
=x+(2x^2+3x-1)/(x^2-2x-3)
=x+(2x^2-4x-6+7x+5)/(x^2-2x-3)
=x+2+(7x+5)/(x^2-2x-3)
=x+2+(7x+5)/(x+1)(x-3)
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