已知x的2次方=x+1,求代数式x的5次方-5x+2的值 详细过程

x²=x+1
以下x²都用x+1代替

平方
x^4=x²+2x+1=x+1+2x+1=3x+2
所以x^5=x(3x+2)=3x²+2x=3(x+1)+2x=5x+3
所以原式=5x+3-5x+2=5

∵x²=x+1 x²-x-1=0 x=(1±√1+4)/2=(1±√5)/2
∴x的5次方-5x+2
=x²*x²*x-5x+2
=(x+1)²*(x+1)²*x-5x+2
=(x²+2x+1)(x²+2x+1)*x-5x+2
=(x+1+2x+1)(x+1+2x+1)*x-5x+2
=(3x+2)²*x-5x+2
=(9x²+12x+4)*x-5x+2
=(9x+9+12x+4)*x-5x+2
=(21x²+13x)-5x+2
=21x+21+12x-5x+2
=28x+23
=28×(1±√5)/2+23
=14±14√5+23
=27±14√5

x²=x+1
x5-5x+2=x(x4-1)+2=x(x²+1)(x²-1)+2=x(x+2)x+2=x²(x+2)+2=(x+1)(x+2)+2=x²+3x+4=4x²+1=
(-1±√3i)²+1=2√3i-1 ……貌似算错了……

x^2=x+1,
x^5-5x+2
=x^5-x-4x+2
=x(x^4-1)-4x+2
=x(x^2+1)(x^2-1)-4x+2
=x(x+2)x-4x+2
=x^2(x+2)-4x+2
=(x+1)(x+2)-4x+2
=x^2+3x+2-4x+2
=x+1-x+4
=5

x²=x+1; x=1+1/x; x-1/x=1
x²-x-1=0
(x-1/x)²=x²+1/x²-2=1
x²+1/x²=3
x^5-5x+2=x^3(x²+1/x²)-6x+2=3x^3-6x+2=3x²(x-1/x)-3x+2=3x²-3x+2=3(x²-x-1)+3+2=5