已知x+4y=-1,xy=5,求(6xy+7y)+【8x-(5xy-y+6x)】的值

来源:学生作业学帮网 编辑:学帮网 时间:2024/04/28 18:50:07

已知x+4y=-1,xy=5,求(6xy+7y)+【8x-(5xy-y+6x)】的值

原式=6xy+7y+8x-5xy+y-6x
=xy+8y+2x
=xy+2(x+4y)
=5+2×(-1)
=3

(6xy+7y)+【8x-(5xy-y+6x)】
=6xy+7y+8x-5xy+y-6x
=xy+8y+2x
=xy+2(x+4y)
=5+2*(-1)
=3

不懂可追问,有帮助请采纳,谢谢!

(6xy+7y)+【8x-(5xy-y+6x)】
=6xy+7y+(8x-5xy+y-6x)
=6xy+7y+2x-5xy+y
=xy+2x+8y
=xy+2(x+4y)
=5+2x(-1)
=5-2
=3

(6xy+7y)+【8x-(5xy-y+6x)】
=6xy+7y+8x-5xy+y-6x
=xy+8y+2x
=xy+2(x+2y)
因:x+4y=-1,xy=5 所以:
原式=5+2x(-1)
=3

x+4y=-1则 x=-1-4y,将x带入xy=5中展开移项得(x+y)(4x+y)=0.解得 y=-1 x=3或y=-4 x=15 ,代入后式得-5或-62