Question

已知x不等于0且x+1除以x=5,求x的4次方+x的四次方分之一

Answer 1

解：x+1/x=5
而x^4+1/x^4=(x^2+1/x^2)^2-2
=[(x+1/x)^2-2]^2-2
=(5^2-2)^2-2
=23^2-2
=529-2
=527

Answer 2

由x+1/x=5
两边平方得：(x+1/x)^2=25
x^2+1/x^2+2=25
x^2+1/x^2=23
两边平方得：（x^2+1/x^2)^2=23^2
x^4+1/x^4+2=529
x^4+1/x^4=527