(1+2x^2)(x-1/x)^8的常数项是什么(求步骤)

问题描述:

(1+2x^2)(x-1/x)^8的常数项是什么(求步骤)
1个回答 分类:数学 2014-09-18

问题解答:

我来补答
(x-1/x)^8
=[(x-1/x)^2]^4
=(x^2-2+1/x^2)^4
=(x^4+4+1/x^4-4x^2-4/x^2+2)^2
=(x^4+1/x^4-4x^2-4/x^2+6)^2
=(x^4+1/x^4)^2+16(x^2+1/x^2)^2+36-2(x^4+1/x^4)(4x^2+4/x^2)-12(4x^2+4/x^2)+12(x^4+1/x^4)
=x^8+2+1/x^8+16x^4+32+16/x^4+36-2(4x^6+4x^2+4/x^2+4/x^6)-48x^2-48/x^2+12x^4+12/x^4
=x^8+2+1/x^8+16x^4+32+16/x^4+36-8x^6-8x^2-8/x^2-8/x^6-48x^2-48/x^2+12x^4+12/x^4
=x^8+1/x^8+28x^4+28/x^4+70-8x^6-56x^2-56/x^2-8/x^6
(1+2x^2)(x-1/x)^8
=(1+2x^2)(x^8+1/x^8+28x^4+28/x^4+70-8x^6-56x^2-56/x^2-8/x^6)
=x^8+1/x^8+28x^4+28/x^4+70-8x^6-56x^2-56/x^2-8/x^6+2x^10+2/x^6+56x^6+56/x^2+140x^2-16x^8-112x^4-112-16/x^4
常数项是70-112=-42
 
 
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