请问:229051=460347x+542990xx+509365xxx+539640xxxx

来源:百度知道 编辑:UC知道 时间:2024/06/07 15:13:19
229051=460347x+542990xx+509365xxx+539640xxxx

请问:以上一元四次方程的x怎么解?x=?
很着急的,谢谢大家帮助。

这么复杂啊,计算机迭代了,x有4个根。你代回方程去试试。
x=
[ -101873/431712+1/16336999252848587041440*56725691850168705005^(1/2)*((-527047964902652647175275*(9520139791010120039639534083238817432304080935858972570607475170620+7722729878112710973010545311708908920060*2885387897601488126652156548849874092986432952215298949^(1/2))^(1/3)+48*(9520139791010120039639534083238817432304080935858972570607475170620+7722729878112710973010545311708908920060*2885387897601488126652156548849874092986432952215298949^(1/2))^(2/3)-20807044982149123986914786611644668204133800640)/(9520139791010120039639534083238817432304080935858972570607475170620+7722729878112710973010545311708908920060*2885387897601488126652156548849874092986432952215298949^(1/2))^(1/3))^(1/2)+1/16336999252848587041440*i*((59794320894652809881611128856727436009502750*(9520139791010120039639534083238817432304080935858972570607475170620+7722729878112710973010545311708908920060*2885387897601488126652156548849874092986432952215298949^(1/2))^(1/3)*((-5270