当x=2008,y=2009时,求代数式(x-y)/x/(x-(2xy-y^2)/x)

题目

当x=2008,y=2009时,求代数式(x-y)/x/(x-(2xy-y^2)/x)

答案

原式=(x-y)/x/((x^2-2xy+y^2)/x)
=(x-y)/x乘以(x/(x^2-2xy+y^2))
=(x-y)/x乘以(x/(x-y)^2)
=1/(x-y)
当x=2008,y=2009
原式=1/(2008-2009)
= -1