Question #038ea

1 Answer
José F.
Oct 28, 2017

There's no trick, just follow the rules

Explanation:

#(x^2+y^2-z^2-2xy)/(x^2-y^2+z^2-2xz)#

#=(x^2-2xy+y^2-z^2)/(x^2-2xz+z^2- y^2)#

#=((x-y)^2-z^2)/((x-z)^2- y^2)#

#=(((x-y)-z)((x-y)+z))/(((x-z)- y)(x-z+ y))#

#=((x-y-z)(x-y+z))/((x-y- z)(x+ y-z))#

#=(x-y+z)/(x+ y-z)#

Related questions
Impact of this question
708 views around the world
You can reuse this answer
Creative Commons License