x与(-y)互为相反数且xy≠0则1910x^2+99y^2/2009xy是多少?
问题描述：
x与(-y)互为相反数且xy≠0则1910x^2+99y^2/2009xy是多少?
答
∵x与(-y)互为相反数
∴x+(-y)=0
∴x=y
(1910x^2+99y^2)/(2009xy)
=(2009x^2)/(2009x^2)
=1
答
因为是相反数
所以设x＝y＝a
则
(1910a^2+99a^2)/-2009a^2
=2009a^2/-2009a^2
=-1