Question

cm rằng \(x^2+y^2+z^2=x.y+y.z+x.z\)

=>x=y=z

Answer 1

Ta có : \(x^2+y^2+z^2=xy+yz+zx\)

   \(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

  \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

 \(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+\left(y^2-2yz+z^2\right)=0\)

 \(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)(1)

\(\text{Mà}\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-z\right)^2\ge0\\\left(y-z\right)^2\ge0\end{cases}}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(2)

\(\text{Từ (1) và (2)}\Rightarrow x-y=y-z=z-x=0\)

                         \(\Rightarrow x=y=z\left(ĐPCM\right)\)