Question

解方程1/(X—3)—1/（X—4）=1/(X—6)-1/(X-7),根据解出的下面方程的解，

猜想方程：1/(x-2006)-1/（x-2007)=1/(x-2009)-1/(x-2010)的解，并验证猜想

Answer 1

1/(x-3)+1/(x-7)=1/(x-4)+1/(x-6)
(x-3+x-7)/(x-3)(x-7)=(x-4+x-6)/(x-4)(x-6)
(2x-10)[1/(x^2-10x+21)-1/(x^2-10x+24)]=0
x^2-10x+21不等于销野x^2-10x+24
所以1/(x^2-10x+21)-1/(x^2-10x+24)不等于0
所以2x-10=0
x=5
经检验，x=5是方程的解
注意到分母分亏携喊别是-3，-4，-6，-7
而解是x=5
所以1/隐滑(x-2006)-1/（x-2007)=1/(x-2009)-1/(x-2010)
则x=2008