若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/28 10:05:43
若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=若有xyz≠0,2y+2z/x=2x+2y/z=2z
若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=
若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=
若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=
将已知变形:2y+2z=kx,2x+2y=kz,2z+2x=ky,三式相加得:4(x+y+z)=k(x+y+z)
即(4-k)(x+y+z)=0,所以:
1) x+y+z≠0时,
2) x+y+z=0时,y+z=-x,代入第一个式子:-2x=kx,因为x≠0,所以:k=-2
综上可知:k=4或-2
若有xyz≠0,2y+2z/x=2x+2y/z=2z+2x/y=k,则K=
若xyz≠0且4(x/z)-5(y/z)=-2,(x/z)+4(y/z)=3,求x:y:z
正数XYZ满足(X+Y)(X+Z)=2则XYZ(X+Y+Z)最大值
若1/2x=1/3y=1/4z,且xyz≠0,求xyz最后一段为“求x:y:z”
已知(x+y)/z=(x+z)/y=(y+z)/x=2,且xyz≠0,则分式(x+y)(x+z)(y+z)/xyz的值为?
x+y+z+2=xyz,x,y,z.为正实数,证明:xyz(x-1)(y-1)(z-1)
(x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z
已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2
若(4x-3y-3z)^2+|x-3y-z|=0 且xyz不等于0 则4x+y-z/x+y+z 的值
若xyz≠0且{4x-5y+2z=0 x+4y-3z=0 ,求x:y:z.
若xyz≠0且4x-5y=-2z,x+4y=3z,求x:y:z
2x+3y+4z,xyz
若实数XYZ满足2|x-y|+√2y+z+z方-z+1//4=0 求X+Y+Z 2y+z在根号里
若x=y/2=z/5,x+y+z=16,xyz=?
已知x+y+z=0,xyz=2,求|x|+|y|+|z|的最小值
xyz>0,x+3y+4z=6.,x^2y^3z最值
若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
设X+Y+Z=0求X^3+X^2Z-XYZ+Y^2Z+Y^3的值