Question

已知x+y分之xy=2

求代数式-x+3xy-y分之-5xy+3x+3y等于？

Answer 1

解：∵xy/(x+y)=2....=>xy=2(x+y)--------①∴(-5xy+3x+3y)/(-x+3xy-y)=[-5xy+3(x+y)]/[-(x+y)+3xy]--------②把①带入②可以得到：[-5X2(x+y)+3(x+y)]/[-(x+y)+3X2X(x+y)]=-7(x+y)/5(x+y)=-7/5

Answer 2

已知x+y分之xy=2，则有xy=2（x+y）-x+3xy-y分之-5xy+3x+3y=-x+3*2(x+y)-y分之-5*2(x+y)+3x+3y =-x+6x+6y-y分之-10x-10y+3x+3y =5x+5y分之-7x-7y接下来就知道怎么做了吧

Answer 3

解：因为(x+y)/xy=2所以x+y=2xy(-x+3xy-y)/(-5xy+3x+3y)=[-(x+y)+3xy)]/[3(x+y)-5xy]将x+y=2xy代入上面式子=(-2xy+3xy)/(3×2xy-5xy)=xy/xy=1