Question

\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{6}\)và\(x^2\).\(y^2\).\(z^2\)=\(288^2\)

Answer 1

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Ta có: \(x^2y^2z^2=288^2\)

\(\Leftrightarrow36k^2=288^2\)

\(\Leftrightarrow k^2=2304\)

Trường hợp 1: k=48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=96\\y=3k=144\\z=6k=288\end{matrix}\right.\)

Trường hợp 2: k=-48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-96\\y=3k=-144\\z=6k=-288\end{matrix}\right.\)

Answer 2

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

\(\Rightarrow4k^29k^236k^2=288^2\)

\(\Rightarrow k^6=64\)

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

\(\Rightarrow\left[{}\begin{matrix}x=4,y=6,z=12\\x=-4,y=-6,z=-12\end{matrix}\right.\)