Question

已知x=2ab/b^2+1,其中a、b都是正数,求[√(a+x)+√(a-x)]/[√(a+x)-√(a-x)]的值

Answer 1

1/x=(b²+1)/(2ab)

a/x=(b²+1)/2b

(a/x)²-1=(b^4+2b²+1)/(4b²)-1

=(b^4-2b²+1)/(4b²)

=(b²-1)²/(4b²)

[√(a+x)+√(a-x)]/[√(a+x)-√(a-x)]

=[√(a+x)+√(a-x)]²/[√(a+x)+√(a-x)][√(a+x)-√(a-x)]

=[a+x+2√(a²-x²)+a-x]/(a+x-a+x)

=[a+√(a²-x²)]/x

=(a/x)+√(a²-x²)/x

=(a/x)+√[(a²-x²)/x²]

=(a/x)+√[(a/x)²-1]

=(b²+1)/2b+√[(b²-1)²/(4b²)]

=(b²+1)/2b+|(b²-1)|/|2b|

b>0

所以

=(b²+1)/2b+|(b²-1)|/2b

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