Question

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

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

Answer 1

解：
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

Answer 2

先化简，变成（x-2xy+y^2）/(x-y)
如果这里没有敲错的话就算吧~如果第一项是x^2的话就容易了
变成（x-y）^2/(x-y)
不用我说了吧

Answer 3

(x-y)/x/(x-(2xy-y^2)/x)=(x-y)/(x^2-2xy+y^2)=(x-y)/(x-y)^2=1/(x-y)=-1

Answer 4

2.77 x 10^（-8）