Question

急：请教一道初二数学题

已知：(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)]/xyz的值
已知：(x+y-z)/z=(x-y+z)/y=(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)]/xyz的值
题目中写错个等号，汗...

题目都写全了，答案是8或-1，不知道怎么算出来的。。。

Answer 1

解：设x+y-z/z=x-y+z/y=y+z-x/x=k

有x+y-z=kz

x-y+z=ky

y+z-x=kx

三式相加得x+y+z=k（x+y+z） k=1

得x+y=（k+1）z

x+z=（k+1）y

y+z=（k+1）x

当x+y+z≠0时

(x+y)(y+z)(x+z)/xyz=（k+1）^3xyz/xyz=(k+1)^3=8

当x+y+z=0时

x+y=-z,x+z=-y,y+z=-x

代入得-1

所以是-1或8

Answer 2

(x+y+z)(x-y+z)(x+y-z)(y+z-x)
=(((x+z)^2-y^2)(y^2-(x-z)^2))
=y^2((x+z)^2+(x-z)^2)-y^4-(x+z)^2(x-z)^2
=y^2(2x^2+2z^2)-y^4-(x^2-z^2)^2
=2x^2y^2+2y^2z^2-y^4-x^4-z^4+2x^2z^2
=2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4

Answer 3

[(x+y)(y+z)(x+z)]/xyz
={[(x+y-z)+z][(y+z-x)+x][(x+z-y)+y]}/xyz
={[(x+y-z)+z]/z}{[(y+z-x)+x]/x}{[(x+z-y)+y]/y}
令(x+y-z)/z=(x-y+z)/y=（y+z-x)/x=K
则[(x+y)(y+z)(x+z)]/xyz=（K+1)^3

不知道你这个K有没有值的！