一道初三一元二次方程题设方程x^2+3x+1=0的两根为X1、X2,求X1^7+X2^7的值

方程x^2+3x+1=0的两根为X1、X2
则有：x1+x2=-3,x1x2=1
x1^2+x2^2=(x1+x2)^2-2x1x2=9-2=7
x1^4+x2^4=(x1^2+x2^2)^2-2x1^2x2^2=49-2*1=47
x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=-3(7-1)=-18
(x1^3+x2^3)(x1^4+x2^4)=-18*47=-846
(x1^3+x2^3)(x1^4+x2^4)=x1^7+x1^3x2^4+x1^4x2^3+x2^7
=(x1^7+x2^7)+x1^3x2^3(x2+x1)
所以：x1^7+x2^7=-846-(x1x2)^3(x2+x1)=-846-1*(-3)=-843

解x^2+3x+1=0,X1=(-3-根号5)/2 X2=(-3+根号5)/2
X1^7+X2^7约=34.172

x^2+3x+1=0
x^2+3x+(3/2)^2-5/4=0
(x+3/2)^2=5/4
x+3/2=±√5/2
x1=-√5/2-3/2
x2=√5/2-3/2
X1^7+X2^7=(-√5/2-3/2)^7+(√5/2-3/2)^7=-843

X1+X2=-3
X1X2=1
X1^2+X2^2=7
X1^3+X2^3=(X1+X2)(X1^2+X2^2-X1X2)=-18 1
X1^4+X2^4=47 2
1式*2式
X1^7+X2^7＋X1^4X2^3＋X1^3X2^4＝-846
X1^7+X2^7=-843