1/(X-3)-1/(X-4)=1/(X-6)-1/(X-7)

来源:百度知道 编辑:UC知道 时间:2024/09/26 20:09:06
根据解出该方程的解,猜想1/(X-2004)-1/(X-2005)=1/(X-2007)-1/(X-2008)
请写出过程

1/(X-3)-1/(X-4)=1/(X-6)-1/(X-7)
-1/(x-3)(x-4)=-1/((x-6)(x-7)
(x-3)(x-4)=(x-6)(x-7)
x^2-7x+12=x^2-13x+42
6x=30
x=5
1/(X-2004)-1/(X-2005)=1/(X-2007)-1/(X-2008)
x^2-4009x+2004*2005=x^2-4015x+2007*2008
6x=12036
x=2006

1/(x-3)+1/(x-7)=1/(x-4)+1/(x-6)
(2x-10)/(x²-10x+21)=(2x-10)/(x²-10x+24)
(2x-10)[1/(x²-10x+21)-1/(x²-10x+24)]=0
因为x²-10x+21≠x²-10x+24
所以1/(x²-10x+21)-1/(x²-10x+24)不等于0
所以2x-10=0
x=5

1/(X-2004)-1/(X-2005)=1/(X-2007)-1/(X-2008)
x^2-4009x+2004*2005=x^2-4015x+2007*2008
6x=12036
x=2006

2006 我是口算了
过程你看下面的吧
呵呵

1/(X-3)-1/(X-4)=1/(X-6)-1/(X-7)
=>1/(X-3(X-4)=1/(X-6)(X-7)
=>(X-3)(X-4)=(X-6)(X-7)
=>x2-7x+12=x2-13x+42
=>6x=30
=>x=5
1/(X-2004)-1/(X-2005)=1/(X-2007)-1/(X-2008)
=>1/(X-2004)(X-2005)=1/(X-2007)(X-200