由(2)得:(x+y)(x-y)-2(x+y)=0,提取公因式:(x+y)(x-y-2)=0
得到x=-y(3),或x=y+2(4)
(3)代入(1)得y^2-y^2+y^2=28,即y=正负2根号7,进而x对应负正2根号7
(4)代入(1)得(y+2)^2+(y+2)y+y^2-28=0
展开解得y=2或y=-4,从而对应x为4或-2
请采纳答案,支持我一下.
由(2)得:(x+y)(x-y)-2(x+y)=0,提取公因式:(x+y)(x-y-2)=0
得到x=-y(3),或x=y+2(4)
(3)代入(1)得y^2-y^2+y^2=28,即y=正负2根号7,进而x对应负正2根号7
(4)代入(1)得(y+2)^2+(y+2)y+y^2-28=0
展开解得y=2或y=-4,从而对应x为4或-2
请采纳答案,支持我一下.