解方程3y+7/y+2+2/2-y=3-（3y/y的平方-4﹚
问题描述：
解方程3y+7/y+2+2/2-y=3-（3y/y的平方-4﹚
答
(3y+7)/(y+2)+2/(2-y)=(3-3y)/(y^2-4)(3y+7)/(y+2)-2/(y-2)=(3-3y)/(y-2)(y+2)(3y+7)(y-2)-2(y+2)=3-3y3y^2+y-14-2y-4=3-3y3y^2-y-18=3-3y3y^2+2y-21=0(3y-7)(y+3)=0y=7/3或y=-3经检验y=7/3或y=-3是方程的解...那个是3减去3y/y的平方-4(3y+7)/(y+2)+2/(2-y)=3-3y/(y^2-4) (3y+7)/(y+2)-2/(y-2)=3-3y/(y-2)(y+2) (3y+7)(y-2)-2(y+2)=3(y-2)(y+2)-3y 3y^2+y-14-2y-4=3y^2-12-3y 3y^2-y-18=3y^2-12-3y -y-18=-12-3y 3y-y=18-12 2y=6 y=3经检验y=3是方程的解