Question

数学问题：已知X²-5X+1=0，求X的4次方减去X的4次方分之一的值？

Answer 1

【参考答案】±115√21

设a、b是方程x²-5x+1=0的两个实数根，
根据韦达定理，a+b=5且ab=1
∴ a>0、b>0
x²-5x=1=0两边同时除以x得：
x+(1/x)=5
∴ x^4 -[1/(x^4)]
=[x²+(1/x²)][x²- (1/x²)]
=[(x+ 1/x)²-2](x+ 1/x)(x- 1/x)
=(5²-2)×5×(x- 1/x)
=115(x- 1/x)
=115×[±√(x- 1/x)²]
=115×{±√[(x+ 1/x)²-4]}
=115×[±√(5²-4)]
=±115√21

Answer 2

X²-5X+1=0
x-5+1/x=0
x+1/x=5
(x+1/x)²=x²+2+1/x²=25
x²+1/x²=23

(x-1/x)²=(x+1/x)²-4=25-4=21
x-1/x=±√21

4次方减去X的4次方分之一
=(x²+1/x²)(x²-1/x²)
=(x²+1/x²)(x+1/x)(x-1/x)
=23×5×√21
=115√21
或
4次方减去X的4次方分之一
=(x²+1/x²)(x²-1/x²)
=(x²+1/x²)(x+1/x)(x-1/x)
=23×5×(-√21)
=-115√21

Answer 3

X²-5X+1=0
x-5+1/x=0
x+1/x=5
(x+1/x)²=5²
x²+2+1/x²=25
x²+1/x²=23
x²-2+1/x²=21
(x-1/x)²=21
x-1/x=±√21

x^4-1/x^4
=(x²+1/x²)(x²-1/x²)
=(x²+1/x²)(x+1/x)(x-1/x)
=23*5*(±√21)
=±115√21