璁綼b涓哄疄鏁颁笖a虏+ab+b虏-a-2b+1=0锛屾眰a+2b镄勫设ab为实数且a²+ab+b²-a-2b+1=0,求a+2b的值已知x=1/2（√2014-√2014分之一）,求（x+√x²+1)²的值

Question

璁綼b涓哄疄鏁颁笖a虏+ab+b虏-a-2b+1=0锛屾眰a+2b镄勫

设ab为实数且a²+ab+b²-a-2b+1=0,求a+2b的值

已知x=1/2（√2014-√2014分之一）,求（x+√x²+1)²的值

 

Answer 1

这是一道很古老的题目了.
算术平方根与平方项均恒非负,两非负项之和=0,两非负项分别=0
a-1=0 a=1
ab-2=0 b=2/a=2/1=2
b=a+1
1/(ab)+1/[(a+1)(b+1)]+...+1/[(a+2013)(b+2013)]
=1/[a(a+1)]+1/[(a+1)(a+2)]+...+1/[(a+2013)(a+2014)]
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+...+1/(a+2013)-1/(a+2014)
=1/a -1/(a+2014)
=1/1-1/(1+2014)
=1-1/2015
=2014/2015