Question

giúp mik vs

Answer 1

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2+2xy=16\)

\(\Leftrightarrow4k^2+2\cdot2k\cdot3k=16\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\\y=3k=3\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-2\\y=3k=-3\end{matrix}\right.\)