y(x+y)-(2x+y)(x-y)-x的2次方,其中x=-2,y=2分之1(先化简,再求值) 先化简,再求值
1个回答
展开全部
这里^2指2次方
1/2为2分之1
y(x+y)-(2x+y)(x-y)-x^2
=xy+y^2-(2x^2-2xy+xy-y^2)-x^2
=xy+y^2-2x^2+xy+y^2-x^2
=2y^2+2xy-2x^2-x^2
=2y^2+2xy-3x^2
当x=-2,y=1/2时
y(x+y)-(2x+y)(x-y)-x^2
=2*(1/2)^2+2*(-2)*(1/2)-3(-2)^2
=1/2-2-12
=13.5
1/2为2分之1
y(x+y)-(2x+y)(x-y)-x^2
=xy+y^2-(2x^2-2xy+xy-y^2)-x^2
=xy+y^2-2x^2+xy+y^2-x^2
=2y^2+2xy-2x^2-x^2
=2y^2+2xy-3x^2
当x=-2,y=1/2时
y(x+y)-(2x+y)(x-y)-x^2
=2*(1/2)^2+2*(-2)*(1/2)-3(-2)^2
=1/2-2-12
=13.5
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询