已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证:x+
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/04/24 02:48:08
已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证:x+y+z=1/x+1/y+1/z.
证明:(x-(yz/x))/(1-yz)=(y-(xz/y))/(1-xz),
十字相乘得:(x-(yz/x))×(1-xz)=(y-(xz/y))×(1-yz),
化简:x-(yz/x)-x²z+yz²=y-(xz/y)-y²z+xz²,
移项:y-x+yz/x-xz/y+x²z-yz²-y²z+xz²=0,
合并同类项(y-x)+z(y/x-x/y)+z(x²-yz-y²+xz)=0,
(y-x)+z(y-x)(y+x)/(xy)+z{(x+y)(x-y)+z(x-y)}=0,
(y-x)(1+z/x+z/y)+z(x-y)(x+y+z)=0,
(y-x){1+z/x+z/y-z(x+y+z)}=0,
因为x≠y,所以1+z/x+z/y-z(x+y+z)=0,即z(x+y+z)=1+z/x+z/y
同÷z,得:x+y+z=1/x+1/y+1/z【证明完毕】
十字相乘得:(x-(yz/x))×(1-xz)=(y-(xz/y))×(1-yz),
化简:x-(yz/x)-x²z+yz²=y-(xz/y)-y²z+xz²,
移项:y-x+yz/x-xz/y+x²z-yz²-y²z+xz²=0,
合并同类项(y-x)+z(y/x-x/y)+z(x²-yz-y²+xz)=0,
(y-x)+z(y-x)(y+x)/(xy)+z{(x+y)(x-y)+z(x-y)}=0,
(y-x)(1+z/x+z/y)+z(x-y)(x+y+z)=0,
(y-x){1+z/x+z/y-z(x+y+z)}=0,
因为x≠y,所以1+z/x+z/y-z(x+y+z)=0,即z(x+y+z)=1+z/x+z/y
同÷z,得:x+y+z=1/x+1/y+1/z【证明完毕】
已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证:x+
设x,y,z∈R+,xy+yz+xz=1,证明不等式:(xy)^2/z+(xz)^2/y+(yz)^2/x+6xyz≥x
XYZ-XY-XZ+X-YZ+Y+Z-1
xyz-xy-xz+x-yz+y+z-1因式分解
已知x,y,z是实数,且xyz=1,求证x^2+y^2+z^2+3大于等于2(xy+xz+yz)
实数xyz=1,求证x^2+y^2+z^2+3>=2(xy+xz+yz)
已知,xyz=0,求x/(xy+x+1)+y/(yz+y+1)+z/(xz+z+1)值?
设x,y,z≥0,且x+y+z=1,求证:0≤xy+yz+xz-2xyz≤7/27
已知 (x²-yz)/x(1-yz)=(y²-xz)/y(1-xz)且x不等于y,x不等于0,y不等
(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)
已知x+y+z=3,xy+yz+xz=-1,xyz=2,求x^2y^2+y^2z^2+x^2z^2
2^x=5^y=10^z(xyz不等于0)求证xy=yz+xz