Question

Tìm x, y, z biết :\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)

Answer 1

Answer 2

Cách 2:

\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)

\(\Rightarrow\dfrac{30x^2}{60}+\dfrac{20y^2}{60}+\dfrac{15z^2}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\)

\(\Rightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\)

\(\Rightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\)

\(\Rightarrow30x^2-12x^2+20y^2-12y^2+15z^2-12z^2=0\)

\(\Rightarrow18x^2+8y^2+3z^2=0\)

Ta có :

\(18x^2\ge0\forall x\) \(;8y^2\ge0\forall y;3z^2\ge0\forall z\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.\Leftrightarrow x=y=z=0\)

Vậy x = y = z =0