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求解高次方程(4次),急x²+(-2x²+√3x)²+(x-√3/2)²+(-2x+√3x)²=3/4

题目详情
求解高次方程(4次),急
x²+(-2x²+√3x)²+(x-√3/2)²+(-2x+√3x)²=3/4
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答案和解析
不要叫我写过程了,我也只能给你答案了,没有必要去计算
总共有4个解,我用matlab算的,只是花点时间抄写而已
x1=0
x2=(3^(1/2)/3 - 2/3)/((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3) + 3^(1/2)/3 + ((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3)
x3= 3^(1/2)/3 - (3^(1/2)/3 - 2/3)/(2*((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3)) - ((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3)/2 - (3^(1/2)*((3^(1/2)/3 - 2/3)/((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3) - ((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3))*i)/2
x4= 3^(1/2)/3 - (3^(1/2)/3 - 2/3)/(2*((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3)) - ((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3)/2 + (3^(1/2)*((3^(1/2)/3 - 2/3)/((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3) - ((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6 + (((17*3^(1/2))/72 + (3^(1/2)*(3^(1/2) - 3))/6)^2 - (3^(1/2)/3 - 2/3)^3)^(1/2))^(1/3))*i)/2