庐山看日出最佳地点:初中数学题

1．≡恒等
2.kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz)
=(kx-y+z)(x+ky-z)(x-y-kz)+(kx-y+z)(x-ky-z)(x-y+kz)
=(kx-y+z)[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)]

(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz)
原式=(kx-y+z)(x+ky-z)(x-y-kz)+(kx-y+z)(x-ky-z)(x-y+kz)
=(kx-y+z)[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)]

然后把[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)]先合并再分解
由于太复杂，我就不便写下去

．≡恒等
2.kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz)
=(kx-y+z)(x+ky-z)(x-y-kz)+(kx-y+z)(x-ky-z)(x-y+kz)
=(kx-y+z)[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)]
同上

1．≡恒等
2.kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz)
=(kx-y+z)(x+ky-z)(x-y-kz)+(kx-y+z)(x-ky-z)(x-y+kz)
=(kx-y+z)[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)]

符号"≡"好象是同余吧
后面的题希望你仔细落实一下"x-ky-z"一项
像这种因式分解题目通常应该是对称的

1恒等
2(kx-y+z)[(x+ky-z)(x-y-kz)+(x-ky-z)(x-y+kz)
第二个提示你(a+b)-(a-b)=a^2-b^2
不想写太多了，自己思考吧