求助一道数学题！求大神指导！

Question

如下：若x^2-3x+1=0，那么x^4+(1/x)^4=?
求详细解答啊！谢谢啦！

Answer 1

由x²-3x+1=0同时除以x可得
x-3+1/x=0
可得 x+1/x=3
又有(x+1/x)²=9=x²+1/x²+2
故x²+1/x²=9-2=7
由((x+1/x)²)²=9²=81=x^4+1/x^4+2+4(x²+1/x²)+4
可得x^4+1/x^4=81-6-4×7
=47

Answer 2

答：
x²+1=3x，两边除以x得x+1/x=3
两边平方得：x²+1/x²+2=9，即x²+1/x²=7
再两边平方得：x^4+1/x^4+2=49
所以x^4+1/x^4=47

Answer 3

x+1/x=3
x^4+(1/x)^4=(x+1/x)^2-2=9-2=7

Answer 4

x^2-3x+1＝0
利用公式法,解得x=3/2±(√5)/2
x+1/x
=3/2±(√5)/2+1/[3/2±(√5)/2]
=3/2±(√5)/2+2/[3/2±(√5)/2]
=3/2±(√5)/2+3/2-+(√5)/2
=3

(x+1/x)^2=9x^2+1/x^2+2=9
x^2+1/x^2=7

(x^2+1/x^2)^2=49
x^4+1/x^4+2=49
x^4+1/x^4=47