Ta có : \(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}\)
\(\Leftrightarrow\frac{30x^2}{60}+\frac{20y^2}{60}+\frac{15z^2}{60}=\frac{12\left(x^2+y^2+z^2\right)}{60}\)
\(\Rightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\)
\(\Leftrightarrow18x^2+8y^2+3z^2=0\)
mà \(18x^2\ge0\) , \(8y^2\ge0\) , \(3z^2\ge0\)
\(\Rightarrow18x^2+8y^2+3z^2\ge0\)
Vậy \(18x^2+8y^2+3z^2=0\Leftrightarrow18x^2=0;8y^2=0;3z^2=0\)
Vậy x = y =z = 0
\(\dfrac{x^{2}}{2}+\dfrac{y^{2}}{3}+\dfrac{z^{2}}{4}=\dfrac{x^{2}+y^{2}+z^{2}}{5}\\\Leftrightarrow \dfrac{30x^{2}+20y^{2}+15z^{2}}{60}=\dfrac{12x^{2}+12y^{2}+12z^{2}}{60}\\\Leftrightarrow 30x^{2}+20y^{2}+15z^{2}=12x^{2}+12y^{2}+12z^{2}\\\Leftrightarrow 18x^{2}+8y^{2}+3z^{2}=0\\18x^{2}\ge0;8y^{2}\ge0;3z^{2}\ge0\\\Rightarrow 18x^{2}+8y^{2}+3z^{52}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=0\)
Vậy \((x;y;z)=(0;0;0)\)