早教吧作业答案频道 -->数学-->
Z^5+Z-1=0求Z的值
题目详情
Z^5+Z-1=0
求Z的值
求Z的值
▼优质解答
答案和解析
五个根分别为
1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) + ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3) - 1/3=0.7549;
1/2 + (3^(1/2)*i)/2=0.5000 + 0.8660i;
1/2 - (3^(1/2)*i)/2=0.5000 - 0.8660i;
- 1/(18*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)/2 - 1/3 + (3^(1/2)*(1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3))*i)/2=-0.8774 - 0.7449i;
- 1/(18*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)/2 - 1/3 - (3^(1/2)*(1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3))*i)/2=-0.8774 + 0.7449i;
1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) + ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3) - 1/3=0.7549;
1/2 + (3^(1/2)*i)/2=0.5000 + 0.8660i;
1/2 - (3^(1/2)*i)/2=0.5000 - 0.8660i;
- 1/(18*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)/2 - 1/3 + (3^(1/2)*(1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3))*i)/2=-0.8774 - 0.7449i;
- 1/(18*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)/2 - 1/3 - (3^(1/2)*(1/(9*((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3)) - ((23^(1/2)*108^(1/2))/108 + 25/54)^(1/3))*i)/2=-0.8774 + 0.7449i;
看了Z^5+Z-1=0求Z的值...的网友还看了以下:
(高分)解一条数学题里的一步,我看不明白是什么,请指出实数x,y和z满足x+y+z=5,xy+yz 2020-05-16 …
甲数是乙数的5分之4,甲数比乙数少百分之【 】,乙数比甲数多百分之【 】甲数比乙数少5分之1,即甲 2020-05-16 …
计算三元一次方程.看一下这道题是不是出错了.{z+y=5,y+z=3,x+z=2.(z+y和y+z 2020-06-03 …
如果O+O=U+U+U,O+Z=U+U+U+U,那么Z+Z+U=()个O.如果设U=6,那么O=( 2020-06-18 …
设方程F(x+z,xy,z)=0确定了隐函数z=z(x,y),其中F具有连续一阶偏导数,求δz/. 2020-06-27 …
高中数学问题急!1、在复平面内,三角形ABC的三个顶点依次对应复数1,2i,5+2i,判断三角形形 2020-07-30 …
下列关于复数的命题命题:1.-a+bi是a+bi的共轭复数2.z1+z2属于R则z1、z2应为共轭 2020-08-01 …
已知复数ω满足ω-4=(3-2ω)i(i为虚数单位),z=5/ω+|z-2|,若z的平方根为a=b 2020-08-02 …
如果(X的平方Y)的A次方乘(XY的B次方Z)的立方乘(Y的平方Z的3立方)2次方,求A,B,C的值 2020-10-31 …
已知(y+z-x)/(x+y+z)=(z+x-y)/(y+z-x)=(x+y-z)/(z+x-y)= 2020-11-01 …