计算：csch{Ln[(1加根号（1226）)/35]}的值

cschx=2/(e^x-1/e^x)

x=ln[(1+√1226)/35]

e^x=(1+√1226)/35

1/e^x=35/(1+√1226)=(√1226-1)/35

e^x-1/e^x=2/35

cschx=2/(2/35)=35

csch{ Ln[ ( 1 + √1226 )/35 ] }

= 2/{ e^Ln[ ( 1 + √1226 )/35 ] - e^-Ln[ ( 1 + √1226 )/35 ] }

= 2/{ e^Ln[ ( 1 + √1226 )/35 ] - 1/e^Ln[ ( 1 + √1226 )/35 ] }

= 2/{ ( 1 + √1226 )/35 - 1/[ ( 1 + √1226 )/35 ] }

= 2/{ ( 1 + √1226 )/35 - 35( 1 - √1226 )/[ ( 1 + √1226 )( 1 - √1226 ) ] }

= 2/{ ( 1 + √1226 )/35 + 35( 1 - √1226 )/1225 }

= 2/{ ( 1 + √1226 )/35 + ( 1 - √1226 )/35 }

= 2 * 35{ 1 + √1226 + 1 - √1226 }

= 70 * 2

= 140 。