
一道数学题(急急急急急)
已知|ab-2|与|b-1|互为相反数,请求代数式:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)的值....
已知|ab-2|与|b-1|互为相反数,请求代数式:
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)的值. 展开
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)的值. 展开
1个回答
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两个绝对值互为相反数,可知他们都=0,可知ab=2,b=1
即a=2,b=1
原式就是
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(3000*3001)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/3000-1/3001)
=1-1/3001
=3000/3001
即a=2,b=1
原式就是
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(3000*3001)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/3000-1/3001)
=1-1/3001
=3000/3001
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