Question

化简:1/(1+√2)+1/(√2+√3)+1/(√3+√4)+1/(√4+√5)+1/(√5+√6)+1/(√6+√7)1/(√7+√8)+1/(√8+√9)

Answer 1

1-√2/(1-√2)(1+√2)+ (√2-√3)/(√2-√3)(√2+√3)+....+(√8-√9)/(√8+√9)(√8-√9)
=-(1-√2)- (√2-√3)-(√3-√4)-....-(√8-√9)
=-1+√9
=2

追问

1-√2/(1-√2)(1+√2)=-(1-√2)怎么算的

追答

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

追问

-(1-√2)- (√2-√3)-(√3-√4)-....-(√8-√9)
=-1+√9怎么算的

Answer 2

∵1／[√n＋√﹙n＋1﹚]=[√﹙n＋1﹚＋√n][√﹙n＋1﹚-√n]／[√n＋√﹙n＋1﹚]＝√﹙n＋1﹚－√n
∴√2－1＋√3－√2＋。。。＋√9－√8＝√9－1＝2