若实数x、y、z满足(x-z)²-4(x-y)(y-z)=0,则下列式子一定成立的是( )A.x+y+z=0 B.x+y-2z=0 C.y+z-2x=0 D.z+x-2y=0
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![若实数x、y、z满足(x-z)²-4(x-y)(y-z)=0,则下列式子一定成立的是( )A.x+y+z=0 B.x+y-2z=0 C.y+z-2x=0 D.z+x-2y=0](/uploads/image/z/2517662-38-2.jpg?t=%E8%8B%A5%E5%AE%9E%E6%95%B0x%E3%80%81y%E3%80%81z%E6%BB%A1%E8%B6%B3%EF%BC%88x-z%EF%BC%89%26%23178%3B-4%EF%BC%88x-y%EF%BC%89%EF%BC%88y-z%EF%BC%89%3D0%2C%E5%88%99%E4%B8%8B%E5%88%97%E5%BC%8F%E5%AD%90%E4%B8%80%E5%AE%9A%E6%88%90%E7%AB%8B%E7%9A%84%E6%98%AF%EF%BC%88+%EF%BC%89A.x%2By%2Bz%3D0+B.x%2By-2z%3D0+C.y%2Bz-2x%3D0+D.z%2Bx-2y%3D0)
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若实数x、y、z满足(x-z)²-4(x-y)(y-z)=0,则下列式子一定成立的是( )A.x+y+z=0 B.x+y-2z=0 C.y+z-2x=0 D.z+x-2y=0
若实数x、y、z满足(x-z)²-4(x-y)(y-z)=0,则下列式子一定成立的是( )
A.x+y+z=0 B.x+y-2z=0 C.y+z-2x=0 D.z+x-2y=0
若实数x、y、z满足(x-z)²-4(x-y)(y-z)=0,则下列式子一定成立的是( )A.x+y+z=0 B.x+y-2z=0 C.y+z-2x=0 D.z+x-2y=0
根据题意
∵(x-z)^2-4(x-y)(y-z)=0,
∴x^2+z^2-2xz-4xy+4xz+4y^2-4yz=0,
∴x^2+z^2+2xz-4xy+4y^2-4yz=0,
∴(x+z)^2-4y(x+z)+4y^2=0,
∴(x+z-2y)^2=0,
∴z+x-2y=0.
所以选D
选D、
(x-z)²=4(x-z)(y-z)≤4*((x-y+y-z)/2)²=(x-z)²
∴只有当等号成立时原式才成立既要满足条件x-y=y-z,这是基本不等式的条件
∴选x+z-2y=0,选D
根据题意
∵(x-z)^2-4(x-y)(y-z)=0,
∴x^2+z^2-2xz-4xy+4xz+4y^2-4yz=0,
∴x^2+z^2+2xz-4xy+4y^2-4yz=0,
∴(x+z)^2-4y(x+z)+4y^2=0,
∴(x+z-2y)^2=0,
∴z+x-2y=0.
所以选D
望采纳
选D
建议选 幺 为最佳。
在解释一下t=x-y,e=y-z,(t+e)^-4te=t^2-2te+e^2=(t-e)^2