在2011.5与它的倒数之间有a个整数,在2011.5与它的相反数之间有b个整数,求(a-b)除以(a+b)的值

kaluosi12022-10-04 11:39:541条回答

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hououjifuu 共回答了20个问题 | 采纳率90%
在2011.5与它的倒数之间有a个整数,2011.5的倒数在0,1之间,那么a=2011.(从1到2011一共2011个整数)
在2011.5与它的相反数之间有b个整数,2011.5的相反数为-2011,.5,那么b=2011*2+1=4023(从-2011到-1有2011个,1到2011有2011个,中间还有一个0)
a-b=-2012,a+b=6034.
(a-b)除以(a+b)的值为1006/3017
1年前

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∵函数f(x)是定义在R上的奇函数,且f(x)的图象关于x=1对称,当0≤x≤1时,f(x)=x,
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f(2011.5)=f(3.5)=f(-0.5)=-f(0.5)=-0.5
故答案为:-0.5

点评:
本题考点: 函数奇偶性的性质.

考点点评: 本题考查了函数的性质,概念,难度不大.

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