初一 数学 数学问题，关于平方的问题 请详细解答,谢谢! (30 11:27:11)

(x+y+z)^2-(-x+y+z)^2+(x-y+z)^2-(x+y-z)^2
=[(y+z)+x]^2-[(y+z)-x]^2+[x-(y-z)]^2-[x+(y-z)]^2
=[(y+z)^2+2x(y+z)+x^2-(y+z)^2+2x(y+z)-x^2]+[x^2-2x(y-z)+(y-z)^2-x^2-2x(y-z)-(y-z)^2]
=4x(y+z)+4x(y-z)
=4x[(y+z)+(y-z)]
=4x*2y
=8xy

或者利用平方差公式：
(x+y+z)^2-(-x+y+z)^2+(x-y+z)^2-(x+y-z)^2
=[(x+y+z)+(-x+y+z)]*[(x+y+z)-(-x+y+z)]+[(x-y+z)+(x+y-z)]*[(x-y+z)-(x+y-z)]
=(2y+2z)*(2x)+(2x)*(2y-2z)
=2x* (2y+2z+2y-2z)
=2x*4y
=8xy

我还回答啊～答案是：4x-4y+4z

（x+y+z）^2—（—x+y+z）^2 +（x—y+z）^2—（x+y—z）^2
=(2y+2z)(2x) + (2x)(2z-2y)
=8xz