Question

(x-y)^2022+(5y-z)^2=0

Answer 1

\(\left(x-y\right)^{2022}+\left(5y-z\right)^2=0\)

Ta thấy: \(\left(x-y\right)^{2022}\ge0\forall x;y\)

              \(\left(5y-z\right)^2\ge0\forall y;z\)

\(\Rightarrow\left(x-y\right)^{2022}+\left(5y-z\right)^2\ge0\forall x;y;z\)

Mặt khác: \(\left(x-y\right)^{2022}+\left(5y-z\right)^2=0\)

nên: \(\left\{{}\begin{matrix}\left(x-y\right)^{2022}=0\\\left(5y-z\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\5y-z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=\dfrac{z}{5}\end{matrix}\right.\)

\(\Leftrightarrow x=y=\dfrac{z}{5}\)

#\(Toru\)