tanh[Ln(根号2加1)]，得出的值

x = ln( √2 + 1 )；

e^x = e^[ ln( √2 + 1 ) ] = √2 + 1；

e^(-x) = 1/e^x = 1/( √2 + 1 ) = ( √2 - 1 )/[ ( √2 + 1 )( √2 - 1 ) ] = √2 - 1；

tanh[ ln( √2 + 1 ) ] = tanhx

= [ e^x - e^(-x) ]/[ e^x + e^(-x) ]

= [ √2 + 1 - ( √2 - 1 ) ]/[ √2 + 1 + ( √2 - 1 ) ]

= [ 2 ]/[ 2√2 ]

= √2/2 。

thx=(e^x-1/e^x)/(e^x+1/e^x)

x=ln(√2+1)

e^x=√2+1=1/(√2-1)

thx=[(√2+1)-(√2-1)]/[(√2+1)+(√2-1)=2/(2√2)=1/√2

或者thx=0.5√2