一个初一的数学题，计算题哈，

来源：百度知道 编辑：UC知道 时间：2024/05/18 08:10:10
若代数式x^2+y^2-14x+2y+50的值为0.
化简求值：[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
请把每一步写清楚了，

如果使代数式x^2+y^2-14x+2y+50的值为0,则
x^2+y^2-14x+2y+50=0
(x-7)^2+(y+1)^2=0
x=7,y=-1

[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
=[2x^2-(x^2-y^2)]-[(x^2-y^2)+2y^2]
=(x^2+y^2)-(x^2+y^2)
=x^2+y^2-x^2-y^2
=0
(好像不应该是这样的呀，原题目中哪儿搞错了？)

您好！

(1)

x^2+y^2-14x+2y+50
=x^2-14x+49+y^2+2y+1
=(x-7)^2+(y+1)^2
x=7
y=-1

(2)

[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
=[2x^2-x^2+y^2]-[x^2-y^2+2y^2]
=2x^2-x^2+y^2-x^2+y^2-2y^2
=0

(1)

x^2+y^2-14x+2y+50
=x^2-14x+49+y^2+2y+1
=(x-7)^2+(y+1)^2
x=7
y=-1

(2)

[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
=[2x^2-x^2+y^2]-[x^2-y^2+2y^2]
=2x^2-x^2+y^2-x^2+y^2-2y^2
=0