Question

设a>0,x=1/2(a1/n-a-1/n),求（x+根号下1+x2）n的值

过程哪！速度点！！！

Answer 1

1+x^2=1+[1/2(a^1/n-a^-1/n)]^2
=[4+a^(2/n)-2+a^(-2/n)]/4
=[a^(2/n)+2+a^(-2/n)]/4
=[1/2(a^1/n+a^(-1/n))]^2
根号(1+x^2)=1/2(a^1/n+a^(-1/n))
(x+根号(1+x^2)^n
=(1/2(a^1/n-a^-1/n)+1/2(a^1/n+a^-1/n))^n
=(a^1/n)^n
=a

Answer 2

1+x^2=1+[1/2(a^1/n-a^-1/n)]^2
=[4+a^(2/n)-2+a^(-2/n)]/4
=[a^(2/n)+2+a^(-2/n)]/4
=[1/2(a^1/n+a^(-1/n))]^2
根号(1+x^2)=1/2(a^1/n+a^(-1/n))

Answer 3

这题我也不会，是单元双侧上的吗，你会吗 ，我们也要做，我也在思考啊