观察下列各式,1/1*2=1-1/2 1/2*3=1/2-1/3,1/3*4=1/3-1/4

用上列变化规律解方程
1/(x(x-1))+1/((x-1)(x-2))+.......+1/((x-9)(x-10))=5/12

化简成1/(x-1)-1/x+1/(x-2)-1/(x-1)+…+1/(x-10)-1/(x-9)=5/12
1/(x-10)-1/x=5/12
10/(x(x-10))=5/12
x=12
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第1个回答  2012-07-16
那下面全化成上面形式 最后x=?咦题目不对啊
第2个回答  2012-07-16
1/(x(x-1))=1/(x-1) -1/x,拆项相销可得方程为1/(x-10)-1/x=10/(x(x-10))=5/12.解得x=12或-2
第3个回答  2012-07-16
同样的都裂开,得1/(x-10)-1/x=5/12,解得x=12。
第4个回答  2012-07-16

第5个回答  2012-07-16
左式=1/x-1/(x-1)+1/(x-1)-1/(x-2)+……+1/(x-9)-1/(x-10)=1/x-1/(x-10),这样就变成了一元二次方程。

观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
化简成1\/(x-1)-1\/x+1\/(x-2)-1\/(x-1)+…+1\/(x-10)-1\/(x-9)=5\/12 1\/(x-10)-1\/x=5\/12 10\/(x(x-10))=5\/12 x=12

观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
原式=1\/1*2+1\/2*3+…+1\/11*12 =1-1\/2+1\/2-1\/3+…+1\/11-1\/12 =1-1\/12 =11\/12

观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
因为1\/n(n+2)=[1\/n-1\/(n+2)]\/2 所以 1\/2*4+1\/4*6+1\/6*8+...+1\/98*100 =1\/2*(1\/2-1\/4)+1\/2*(1\/4-1\/6)+1\/2*(1\/6-1\/8)+...+1\/2*(1\/98-1\/100)=1\/2*(1\/2-1\/4+1\/4-1\/6+1\/6-1\/8+...+1\/98-1\/100)=1\/2*(1\/2-1\/100)=49\/200.

...等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边...
观察下列等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边分别相加得:1\/1*2+1\/2*3+1\/3*4=1-1\/2+1\/2-1\/3+1\/3-1\/4=1-1\/4=3\/4.(1)猜想并写出:1\/n(n+1)=1\/n-1\/(n+1);(2)直接写出下列各式的计算结果 1\/1*2+1\/2*3+1\/3*4+...

观察下列各式:1\/1×2=1-1\/2;1\/2×3=1\/2-1\/3;1\/3×4=1\/3-1\/4,···
解:(1) 1\/n(n+1)=1\/n-1\/(n+1)(2)1\/2+1\/6+1\/12+···+1\/24=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(15-1\/16)=1-1\/2+1\/2-1\/3...+1\/15-1\/16 =1-1\/16 =15\/16

观察下列等式:1\/1x2=1-1\/2,1\/2x3=1\/2-1\/3,1\/3x4=1\/3-1\/4
1\/n-1\/(n+1)2009\/2010

...等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边...
解:这是分式相消的思想!观察结果:注意:1\/1*2 = (2-1)\/1*2 = 2\/1*2 - 1\/1*2 = 1\/1 - 1\/2 1\/2*3 = (3-2)\/2*3 = 3\/2*3 - 2\/2*3 = 1\/2 - 1\/3 1\/3*4 = (4-3)\/3*4 = 4\/3*4 - 3\/3*4 = 1\/3 - 1\/4 上述各式相加:左边=1\/1*2+1\/2*3+...

观察下列各式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3.1\/3*4=1\/3-1\/4,……
①1\/8*9=1\/8-1\/9 ②1\/n(n+1)=1\/n-1\/(n+1) (n是正整数)(2)由以上的几个式子及你所找到的规律计算:1\/1*2+1\/2*3+1\/3*4+…+1*2010*2011 =1-1\/2011 =2010\/2011

观察下列各式1\/1x2=1-1\/2,1\/2x3=1\/2-1\/3,1\/3x4=1\/3-1\/4……
(1)根据以上的式子填写下列各题 ①1\/9×10=( 1\/9 -1\/10 )②1\/n(n+1)=[ 1\/n -1\/(n+1) ] n是正整数 (2)由以上几个式子能找到的规律计算:1\/1×2+1\/2×3+1\/3×4+……+1\/2011×2012 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/2011-1\/2012 =1-1\/2012 =2011\/2012 ...

观察下列等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4?
所以,原式=(1\/2-1\/4+1\/4-1\/6+1\/6-1\/8+…+1\/2006-1\/2008)\/2 =(1\/2-1\/2008)\/2=1003\/4016,2,。。。,2,观察下列等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边分别相加得:1\/1*2+1\/2*3+1\/3*4=1-1\/2+1\/2-1\/3+1\/3-1\/4=1...

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