已知x+3y=2,求代数式(x²-y²/x+y)-【2(x+y)】的值
问题描述：
已知x+3y=2,求代数式(x²-y²/x+y)-【2(x+y)】的值
答
已知x+3y=2
则(x²-y²)/(x+y)-2(x+y)
=x-y-2x-2y
=-x-3y
=-(x+3y)
=-2
如果不懂，请追问，祝学习愉快！
答
原式=(x+y)(x-y)/(x+y)-2x-2y
=x-y-2x-2y
=-x-3y
=-(x+3y)
=-2
答
原式=[(x+y)(x-y)/(x+y)]-2(x+y)
=x-y-(2x+2y)
=x-y-2x-2y
=-x-3y
=-(x+3y)
=-2