Question

已知x的平方＝x＋1求代数式x的五次方－5x＋2的值？

Answer 1

x²=x+1
两边平方
x^4=x²+2x+1=(x+1)+2x+1=3x+2
x^5=x×x^4=x(3x+2)
=3x²+2x
=3(x+1)+2x
=5x+3
所以原式=(5x+3)-5x+2
=5,1,x^2 = x+1
x^5 - 5x +2 = (x^2)^2 * x -5x +2
= [(x+1)^2] *x -5x +2 ( x^2 用x+1代替）
=x^3 + 2x^2 +x -5x +2
=(x^2)*x +2x^2-4x +2
=(x+1)*x +2x^2 -4x +2
=3x^2 - 3x +2
= 3(x+1) -3x +2
=5,2,x^5-5x+2
=x(x^2)^2-5x+2
=x(x+1)^2-5x+2
=x(x^2+2x+1)-5x+2
=x(x+1+2x+1)-5x+2
=x(3x+2)-5x+2
=3x^2-3x+2
=3(x+1)-3x+2=5,1,x5-5x+2
=x(x4-4)-x+2
=x(x2-2)(x2+2)-x+2 代入x2=x+1
=x(x+1-2)(x+1+2)-x+2
=x(x-1)(x+3)-x+2
=(x2-x)(x+3)-x+2 代入x2=x+1
=(x+1-x)(x+3)-x+2
=x+3-x+2
=3+2
=5,1,