Question

先化简，再求值：（x-y分之一+x+y分之一）÷x的平方-y的平方分之x的平方y，其中x=根号3+1，y=根号3-1

Answer 1

x=√3+1，y=√3-1
[1/(x-y)+1/(x+y)] ÷ x^2y/(x^2-y^2)
= (x+y+x-y)/[(x+y)(x-y) × (x+y)(x-y)/(x^2y)
= 2x/(x^2y)
= 2/(xy)
= 2/[(√3+1)(√3-1)]
= 2/(3-1)
= 1

Answer 2

[1/(x-y)+1/(x+y)]÷(x^2y)/(x^2-y^2)
=[(x+y+x-y)/(x-y)(x+y)]÷(x^2y)/(x+y)(x-y)
=[2x/(x-y)(x+y)]÷(x^2y)/(x+y)(x-y)
=2x/(x^2y)
=2/(xy)
=2/[(√3+1)×(√3-1)]
=2/2
=1