a、b满足√（a-1)+√(b-2)=0,
则
a-1=0,b-2=0
a=1,b=2
所以
1／ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
=1/1×2+1/2×3+.+1/2002×2003
=1-1/2+1/2-1/3+.+1/2002-1/2003
=1-1/2003
=2002/2003
已知字母ab满足已知：字母a、b满足√（a-1)+√(b-2)=0,求1／ab+1/(a+1)(b+1)+1/(a+2)
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