临川唱凯中学校长:我要问一道初二的奥数题

来源:百度文库 编辑:杭州交通信息网 时间:2024/05/04 14:43:41
已知(Z-X)的平方-4(X-Y)(Y-Z)求证:2Y=X+Z我在此先谢谢了,请哪位高手帮一下,我想问一下下标和平方怎么打啊,还有轮换对称式分解有什么技巧啊,分解因式(X+Y)(Y+Z)(Z+X)+XYZ,谢谢啊

(Z-X)的平方=4(X-Y)(Y-Z)
[(x-y)+(y-z)]^2=4(x-y)(y-z)
(x-y)^2+(y-z)^2-2(x-y)(y-z)=0
[(x-y)-(y-z)]^2=0
所以x-y=y-z
2y=x+z

轮换对称式将一个作为主要未知数
(X+Y)(Y+Z)(Z+X)+XYZ
=[x^2+(y+z)x+yz](y+z)+xyz
=(y+z)x^2+(y+z)^2x+yz(y+z)+xyz
=(y+z)x^2+[(y+z)^2+yz]x+yz(y+z)

y+z yz
1 y+z

(X+Y)(Y+Z)(Z+X)+XYZ
=(x+y+z)[(y+z)x+yz]
=(x+y+z)(xy+yz+zx)

打平方很简单,在word中就可以了
解:原式分解可得
=3xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y
=(x+y+z)yz+(x+y+z)xz+(x+y+z)xy
=(x+y+z)(xy+yz+xz)