2007-x/2009 2009-x/2011=2011-x/2013 2013-x/2015

问题描述:

2007-x/2009 2009-x/2011=2011-x/2013 2013-x/2015
尽快↖(^ω^)↗↖(^ω^)↗

(2007-x)/2009+(2009-x)/2011=(2011-x)/2013+(2013-x)/2015
(2009-2-x)/2009+(2011-2-x)/2011=(2013-2-x)/2013+(2015-2-x)/2015
1-(x+2)/2009+1-(x+2)/2011=1-(x+2)/2013+1-(x+2)/2015
(x+2)/2009+(x+2)/2011=(x+2)/2013+(x+2)/2015
(x+2)[1/2009+1/2011-1/2013-1/2015)=0
x+2=0
x=-2