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两道多项式计算
(a+2b)(a-2b)[(a+2b)^2-6ab][(a-2b)^2+6ab]
(a-1)(a^4+a^3+a^2+a+1)
前面那题我钻研出来了,方法和你一样,后面的我就卡在(a-1)[a^3(a+1)+a(a+1)+1)]
这一步上了,我去验证一下……再回来给你加最佳……
人气:489 ℃ 时间:2020-03-30 14:14:03
解答
(a+2b)(a-2b)[(a+2b)^2-6ab][(a-2b)^2+6ab]
=(a+2b)(a-2b)[a^2+4b^2-2ab][a^2+4b^2+2ab]
=(a^2-4b^2)[(a^2+4b^2)^2-4a^2b^2]
=(a^2-4b^2)*(a^2+4b^2)^2-4a^2b^2(a^2-4b^2)
=a^6+4a^4b^2-16a^2b^4-64b^6-4a^4b^2+16a^2b^4
=a^6-64b^6
(a-1)(a^4+a^3+a^2+a+1)=(a-1)[a^3(a+1)+a(a+1)+1)]
=(a-1)*a^3(a+1)+(a-1)*a(a+1)+a-1
=a^3(a^2-1)+a(a^2-1)+a-1
=a^5-a^3+a^3-a+a-1
=a^5-1
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