已知√(a-1)+(ab-2)²=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
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已知√(a-1)+(ab-2)²=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
已知√(a-1)+(ab-2)²=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
已知√(a-1)+(ab-2)²=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值
√(a-1)+(ab-2)²=0
则必有 A-1=0 AB-2=0 ==>A=1 B=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)
=1/2+1/(2*3)+1/(3*4)+.+1/(2009*2010)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2009-1/2010)
=1-1/2010=2009/2010
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