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!

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...