Question

已知x2-x-1=0，求x5-x4-3x3+3x2+x的值

Answer 1

x5-x4-3x3+3x2+x
=x3(x2-x-3)+3x2+x
=x3(x2-x-1-2)+3x2+x
x2-x-1=0带入上式
原式=-2x3+3x2+x
=-x(2x2-2x-x-2+1)
=-x(2(x2-x-1)-x+1))
x2-x-1=0带入上式
原式=-x*(-x+1)=x2-x=1

Answer 2

最终的结果是1，用降幂法解决

Answer 3

【】【】【】
x2-x-1=0，
求x5-x4-3x3+3x2+x
【因为】x2-x=1
x5-x4-3x3+3x2+x
=x^3(x^2-x)-3x(x^2-x)+x
=x^3-2x
=x^3-x^2+x^2-x-x
=x+1-x
=1

Answer 4

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

Answer 5