编写一道应用题,使其中的未知数x满足分式方程20/3+x=15/x恩恩。那个还有1~小小道。(6y+12/y^2+4y+4)-(y^2-4/y^2-4y+4)+(y^2/y^2-4)=0

1.乙每小时比甲多走20/3千米,两人的步速之积为15千米/小时,求甲的步速
2.(6y+12)/(y^2+4y+4)-(y^2-4)/(y^2-4y+4)+y^2/(y^2-4)
=6(y+2)/(y+2)^2-(y+2)(y-2)/(y-2)^2+y^2/[(y+2)(y-2)]
=6/(y+2)-(y+2)/(y-2)+y^2/[(y+2)(y-2)]
=[6(y-2)-(y+2)^2+y^2]/[(y+2)(y-2)]
=(2y-16)/[(y+2)(y-2)]
=(2y-16)/(y^2-4)