d: Ta có: 5x=2y
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
Ta có: \(x^3+y^3=133\)
\(\Leftrightarrow k^3=1\)
hay k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\\y=5k=5\end{matrix}\right.\)


