Question

已知x(x-1)-(x的2次方-y)=-2,求（x的2次方+y的2次方/2）-xy

Answer 1

此题应为：已知x（x-1)-(x^2-y)= -2 求（x^2+y^2)/2-xy的值。
解：由 x(x-1)-(x^2-y)= -2
展开 得 x^2-x-x^2+y= -2
x-y=2
所以 (x^2+y^2)/2-xy
=x^2/2+y^2/2-2xy/2
=(x-y)^2/2
将x-y=2整体代入
得 =(x-y)^2/2
=4/2
=2

Answer 2

此题表述有错误，应为：已知x（x-1)-(x^2-y)= -2 求（x^2+y^2)/2-xy的值。
解：x(x-1)-(x^2-y)= -2
x^2-x-x^2+y= -2
x-y=2
则(x^2+y^2)/2-xy
=x^2/2+y^2/2-2xy/2
=(x-y)^2/2
=4/2
=2

Answer 3

（x的2次方+y的2次方/2）中的/2是否包括x的2次方

Answer 4

x(x-1)-(x^2-y)=-2
x^2-x-x^2+y=-2
x-y=2
(x^2+y^2)/2-xy
=[(x-y)^2]/2
=2