答：
(x+y+z-2)^2+(y+z+2-x)^2
=(x+y+z-2+y+z+2-x)^2-2(x+y+z-2)(y+z+2-x)
=(2y+2z)^2-2*[(y+z)^2-(2-x)^2]
=4(y+z)^2-2(y+z)^2+2(x-2)^2
=2(y+z)^2+2(x-2)^2可以告訴我原理么= =謝謝！先用和的平方公式：(a+b)^2=a^2+2ab+b^2后用平方差公式：c^2-d^2=(c+d)(c-d)2(y+z)^2+2(x-2)^2=2y^2+4yz+2z^2+2x^2-4x+8結果是一樣的