Question

已知x的平方-x-1=0,求x的5次方-5x-3的值,要有具体的过程？

Answer 1

以下用 x^n 表示 x的n次方.
= = = = = = = = =
因为 x^2 -x -1=0,
所以 x^2 =x +1.
所以 x^3 =x (x+1)
=x^2 +x
=(x+1)+x
=2x +1,
x^4 =x (2x +1)
=2x^2 +x
=2(x+1) +x
=3x +2,
x^5 =x (3x +2)
=3x^2 +2x
=3(x+1) +2x
=5x +3.
所以 x^5 -5x -3 =0.
= = = = = = = = =
降幂.
x^2 =x+1 是关键.
然后再分别求 x^3,x^4,x^5.
再高的次数也不怕.,9,x的平方-x-1=0,x的平方=x+1.x的五次方-5x-3=x乘（x的四次方-5）-3=x乘【(x的平方)的平方-5】-3=x乘【（x+1)的平方-5】-3=x乘（x的平方+2x+1-5）-3=x乘（x+1+2x+1-5）-3=3x的平方-3x-3=3(x+1)-3x-3=0,2,由x2-x-1=o 得x2=x+1
X5-5x-3=x2*x3-5x-3=(x+1)x3-5x-3=x4+x3-5x-3=(x+1)2+(x+1)x-5x-3=x2+2x+1+x2+x-5x-3=2x2-2x-2=2(x2-x-1)=0,1,x^2-x-1=0
∴x^5-5x-3
=(x^5-x^4-x^3)+x^4+x^3-5x-3
=x^3(x^2-x-1)+x^4+x^3-5x-3
=x^4+x^3-5x-3
=(x^4-x^3-x^2)+2x^3+x^2-5x-3
=2x^3+x^2-5x-3
=2(x^3-x^2-x)+3x^2-3x-3
=3x^2-3x-3
=3(x^2-x-1)
=0,0,