关于分式的加减.已知:1/x(x+1)=(1/x - 1/x+1)*1,1/x(x+2)=(1/x - 1/x+2)*1/2,1/x(x+3)=(1/x - 1/x+3)*1/3,…………1/x(x+10)=(1/x - 1/x+10)*1/10则:1/(x+m)(x+n)=_________________;探究:若|ab-2|+(b-1)²=0,试求1/ab + 1/(a+1)(b+
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![关于分式的加减.已知:1/x(x+1)=(1/x - 1/x+1)*1,1/x(x+2)=(1/x - 1/x+2)*1/2,1/x(x+3)=(1/x - 1/x+3)*1/3,…………1/x(x+10)=(1/x - 1/x+10)*1/10则:1/(x+m)(x+n)=_________________;探究:若|ab-2|+(b-1)²=0,试求1/ab + 1/(a+1)(b+](/uploads/image/z/2765743-7-3.jpg?t=%E5%85%B3%E4%BA%8E%E5%88%86%E5%BC%8F%E7%9A%84%E5%8A%A0%E5%87%8F.%E5%B7%B2%E7%9F%A5%EF%BC%9A1%2Fx%28x%2B1%29%3D%281%2Fx+-+1%2Fx%2B1%29%2A1%2C1%2Fx%28x%2B2%29%3D%281%2Fx+-+1%2Fx%2B2%29%2A1%2F2%2C1%2Fx%28x%2B3%29%3D%281%2Fx+-+1%2Fx%2B3%29%2A1%2F3%2C%E2%80%A6%E2%80%A6%E2%80%A6%E2%80%A61%2Fx%28x%2B10%29%3D%281%2Fx+-+1%2Fx%2B10%29%2A1%2F10%E5%88%99%EF%BC%9A1%2F%EF%BC%88x%2Bm%29%28x%2Bn%EF%BC%89%3D_________________%EF%BC%9B%E6%8E%A2%E7%A9%B6%EF%BC%9A%E8%8B%A5%7Cab-2%7C%2B%28b-1%29%26sup2%3B%3D0%2C%E8%AF%95%E6%B1%821%2Fab+%2B+1%2F%28a%2B1%29%28b%2B)
关于分式的加减.已知:1/x(x+1)=(1/x - 1/x+1)*1,1/x(x+2)=(1/x - 1/x+2)*1/2,1/x(x+3)=(1/x - 1/x+3)*1/3,…………1/x(x+10)=(1/x - 1/x+10)*1/10则:1/(x+m)(x+n)=_________________;探究:若|ab-2|+(b-1)²=0,试求1/ab + 1/(a+1)(b+
关于分式的加减.
已知:1/x(x+1)=(1/x - 1/x+1)*1,
1/x(x+2)=(1/x - 1/x+2)*1/2,
1/x(x+3)=(1/x - 1/x+3)*1/3,
…………
1/x(x+10)=(1/x - 1/x+10)*1/10
则:1/(x+m)(x+n)=_________________;
探究:若|ab-2|+(b-1)²=0,试求1/ab + 1/(a+1)(b+1) + 1/(a+2)(b+2) + 1/(a+3)(b+3) +……+ 1/(a+2007)(b+2007) + 1/(a+2008)(b+2008) 的值.
关于分式的加减.已知:1/x(x+1)=(1/x - 1/x+1)*1,1/x(x+2)=(1/x - 1/x+2)*1/2,1/x(x+3)=(1/x - 1/x+3)*1/3,…………1/x(x+10)=(1/x - 1/x+10)*1/10则:1/(x+m)(x+n)=_________________;探究:若|ab-2|+(b-1)²=0,试求1/ab + 1/(a+1)(b+
1/(x+m)(x+n)=[1/(x+m)-1/(x+n)]*1/(n-m)
|ab-2|+(b-1)²=0
ab-2=0,b-1=0
b=1
a=2
1/ab + 1/(a+1)(b+1) + 1/(a+2)(b+2) + 1/(a+3)(b+3) +……+ 1/(a+2007)(b+2007) + 1/(a+2008)(b+2008)
=1/2+1/6+1/12+1/20+……+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010
=1-1/2010
=2009/2010
第一题是 |1/(x+m) - 1/(x+n)| * |m-n| 这个你把答案通分一下就知道跟那个乘积一样了
第二题是,如果满足上面那个式子=0,必须绝对值里的东西和平方里的东西都=0,因为这两个东西都是非负的,所以a=2,b=1
于是要求的式子变成:
1/1*2 + 1/2*3 + 1*3*4 + ... + 1/2009*2010
= (1/1 - 1/2) +...
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第一题是 |1/(x+m) - 1/(x+n)| * |m-n| 这个你把答案通分一下就知道跟那个乘积一样了
第二题是,如果满足上面那个式子=0,必须绝对值里的东西和平方里的东西都=0,因为这两个东西都是非负的,所以a=2,b=1
于是要求的式子变成:
1/1*2 + 1/2*3 + 1*3*4 + ... + 1/2009*2010
= (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/2009 - 1/2010) (中间的都抵消了)
=1-1/2010
=2009/2010
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