题目

解方程组 4x^2-y=0 3x^2+xy+x-2y+6=0

提问时间：2021-03-25

答案

x1 = -1/12*(1261+234*29^(1/2))^(1/3)-13/12/(1261+234*29^(1/2))^(1/3)+5/12
x2 = 1/24*(1261+234*29^(1/2))^(1/3)+13/24/(1261+234*29^(1/2))^(1/3)+5/12+1/2*i*3^(1/2)*(-1/12*(1261+234*29^(1/2))^(1/3)+13/12/(1261+234*29^(1/2))^(1/3))
x3 = 1/24*(1261+234*29^(1/2))^(1/3)+13/24/(1261+234*29^(1/2))^(1/3)+5/12-1/2*i*3^(1/2)*(-1/12*(1261+234*29^(1/2))^(1/3)+13/12/(1261+234*29^(1/2))^(1/3))
x1 = -0.7971
y1 = 2.5416
x2 = 1.0236 - 0.9133i
y2 = 0.8546 - 7.4788i
x3 = 1.0236 + 0.9133i
y3 = 0.8546 + 7.4788i