已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值

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已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值已知(a-1)^2+│ab-2│=0求1/ab+1/

已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值

已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
因为平方和绝对值都是非负的,已知中两者相加为0,只能是两者都为0
于是 a=1,b=2
同时1/ab=1/a-1/b=1-1/2
1/(a+1)(b+1)=1/(2*3)=1/2-1/3
...
1/(a+2005)(b+2005)=1/(2006*2007)=1/2006-1/2007
累加可得
等式右边的被减数与下一等式右边的减数抵消
故结果为
=1-1/2007=2006/2007