Question

已知x+x的负一次方=3，求x的平方减x的负二次方的值

Answer 1

X的负一次方也就是X分之一,故:
x+1/x=3
两边平方并展开整理得:
x^2+1/x^2=7
(x-1/x)^2=x^2+1/x^2-2=7-2=5
x-1/x=±√5
故x^2-1/x^2=(x+1/x)(x-1/x)=3(x-1/x)=3*（±√5）=±3√5

Answer 2

∵x+1/x=3
∴（x+1/x)²=9
x²+2+1/x²=9
x²-2+1/x²=5
(x-1/x)²=5
∴x-1/x=±√5
∴x²-1/x²
=(x+1/x)(x-1/x)
=3×(±√5)
=±3√5
即x^2-x^(-2)=±3√5

Answer 3

x+1/x=3
x²+1/x²+2=9
x²+1/x²=7
x²+1/x²-2=5
(x-1/x)²=5
x-1/x=±√5
x²-1/x²=(x+1/x)(x-1/x)=±3√5

Answer 4

（x+1/x)²=9
=> x²+1/x²=7
=>x4+1/x4=47
=>x4+1/x4-2=45
=>(x²+1/x²)²=45
=>x²+1/x²=3√5