第8题求数学学霸解答谢谢~
解:令(x+y-2z)/z=(x-2y+z)/y=(-2x+y+z)/x=k
则x+y-2z=kz
x+z-2y=ky
y+z-2x=kx
等式左右两边相加,得
2(x+y+z)-2(x+y+z)=k(x+y+z)
即k(x+y+z)=0
则x+y+z=0或k=0
当x+y+z=0时,原式=-1
当k=0时,因为xyz≠0,
所以x+y-2z=0,x-2y+z=0,-2x+y+z=0
即x+y=2z,x+z=2y,y+z=2x
所以原式=8
综上所述,原式等于-1或8
(x+y-2z)/z=(x-2y+z)/y=(-2x+y+z)/x
解:∵分母之和为x+y+z
①当x+y+z=0时,x+y=-z,y+z=-x,z+x=-y
∴(x+y)(y+z)(z+x)/xyz=-xyz/(xyz)=-1
②当x+y+z≠0时,(xyz≠0)
(x+y-2z)/z=(x-2y+z)/y=(-2x+y+z)/x=[(x+y-2z)+(x+y-2z)+(x+y-2z)]/(x+y+z)=0
∴x+y-2z=0,x-2y+z=0,-2x+y+z=0
∴x+y=2z,y+z=2x,x+z=2y
∴(x+y)(y+z)(z+x)/xyz=2x*2y*2z/(xyz)=8
即:答案为-1或8
(x+y-2z)/z=(x-2y+z)/y
(x+y)/z-2=(x+z)/y-2
(x+y)/z=(x+z)/y
y(x+y)=z(x+z)
xy+y^2=xz+z^2
(y^2-z^2)+(xy-xz)=0
(y+z)(y-z)+x(y-z)=0
(x+y+z)(y-z)=0
同理
由(x+y-2z)/z=(-2x+y+z)/x,可得(x+y+z)(z-x)=0
由(x-2y+z)/y=(-2x+y+z)/x,可得(x+y+z)(x-y)=0
当x+y+z=0时,x+y=-z, x+z=-y, y+z=-x
所以(x+y)(y+z)(z+x)/xyz=(x+y)/z*(y+z)/x*(z+x)/y=(-1)*(-1)*(-1)=-1
当x+y+z≠0时,那么x=y=z
所以
(x+y)(y+z)(z+x)/xyz=2x*2y*2z/xyz=8