(√2+1)(√2-1)=1;(√3+√2)(√3-√2)=1;(√4+√3)(√4-√3)=1 观察上面的规律,计算下列式子的值

问题描述:

(√2+1)(√2-1)=1;(√3+√2)(√3-√2)=1;(√4+√3)(√4-√3)=1 观察上面的规律,计算下列式子的值
(√2+1)(√2-1)=1;(√3+√2)(√3-√2)=1;(√4+√3)(√4-√3)=1
观察上面的规律,计算下列式子的值
:【1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(√2012+√2011)】(√2012+1)=?

:【1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(√2012+√2011)】(√2012+1)
=(√2-1+√3-√2+√4-√3+……+√2012-√2011)×(√2012+1)
=(√2012-1)(√2012+1)
=2012-1
=2011