Question

已知,x^2-4x+1=0 求x^2+1/x^2 x^3+1/x^3 x^5+1/x^5

Answer 1

x^2-4x+1=0
x-4+1/x=0 两边除以x
∴x+1/x=4
两边平方得
(x+1/x)²=16
x²+2+1/x²=16
∴x²+1/x²=14

x^3+1/x^3=(x+1/x)(x²-1+1/x²)=4×(14-1)=52
(x²+1/x²)(x^3+1/x^3)=x^5+1/x+x+1/x^5=x^5+1/x^5+(x+1/x)
∴14×52=x^5+1/x^5+4
x^5+1/x^5=724

Answer 2

x^2-4x+1=0
显然x≠0
两边同时除以x，得到x+（1/x）=4
x^2+1/x^2=[x+（1/x）]^2 -2=14
x^3+1/x^3=(x^2+1/x^2)[x+（1/x）]-(x+(1/x))=14*4-4=42
x^5+1/x^5=（x^3+1/x^3）（x^2+1/x^2）-（x+1/x）=42*14-4=584