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【谜语一则】猜猜猜……

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发表于 2016-9-9 21:33 | 显示全部楼层 |阅读模式
本帖最后由 风花飘飘 于 2016-9-10 08:42 编辑

带着安全套拜佛。
答:成语一。

今有方程X^5-5*X-2=0

X^5-5*X-2分解因式:
=(X^3+R*X^2+S*X-(2/(-4)^(1/5)))*(X^2-R*X+(-4)^(1/5))

=(X^3+R*X^2+(((4^(1/5)*R^2+16^(1/5))*X^2+(16^(1/5)*R-2)*X+2*R-5*4^(1/5))/(4^(1/5)*X^2-4^(1/5)*R*X-16^(1/5)))*X-(2/(-4)^(1/5)))*(X^2-R*X+(-4)^(1/5))


或者:
= (X^3+R*X^2+(((4^(1/5)*R^2+16^(1/5))*X^2+(16^(1/5)*R-2)*X+2*R-5*4^(1/5))/(4^(1/5)*X^2-4^(1/5)*R*X-16^(1/5)))*X+(2/(4)^(1/5)))*(X^2-R*X-(4)^(1/5))
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