若x/2=y/3,则分式7x^2-y^2/x^2-2xy+3y^2的值是 多少?
展开全部
解：由已知得：x=2y/3，代入欲求式子，得
(7x^2-y^2)/(x^2-2xy+3y^2)
=[7*(2y/3)^2-y^2]/[(2y/3)^2-2*2y/3*y+3y^2]
=(28y^2/9-y^2)/(4y^2/9-4y^2/3+3y^2)
=(19y^2/9)/(19y^2/9)
=1
(7x^2-y^2)/(x^2-2xy+3y^2)
=[7*(2y/3)^2-y^2]/[(2y/3)^2-2*2y/3*y+3y^2]
=(28y^2/9-y^2)/(4y^2/9-4y^2/3+3y^2)
=(19y^2/9)/(19y^2/9)
=1
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
