题目

数学三次一元方程和四次一元方程解答
+ 7X^2 + 12 = 0
X^6 - 7X^3 - 8 = 0
（X+4）（X^2-7x+49）= 5
X^3 + 12 = 3X^2 + 4X
X^3 - 5.3X^2 +1.46X + 0.2 = 0 (其中X1=5)

提问时间：2020-10-30

答案

X^4+7X²+12=0
(x²+3)(x²+4)=0
x1=√3i,x2=-√3i,x3=2i,x4=-2i
X^6-7X³-8=0
(x³-8)(x³+1)=0
x1=-1,x2=2
X³+12=3X²+4X
x³-4x=3x²-12
x(x+2)(x-2)=3(x+2)(x-2)
(x-3)(x-2)(x+2)=0
x1=3,x2=2,x3=-2
X³- 5.3X²+1.46X+0.2=0
x³-5x²-0.3x²+1.5x-0.04x+0.2=0
x²(x-5)-0.3x(x-5)-0.04(x-5)=0
(x²-0.3x-0.04)(x-5)=0
(x+0.1)(x-0.4)(x-5)=0
x1=-0.1,x2=0.4,x3=5