Question

Answer 1

\(x:y:z=3:4:5=>\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

\(=>\left\{{}\begin{matrix}x=\dfrac{3y}{4}\\z=\dfrac{5y}{4}\end{matrix}\right.\) thế x,z vào \(2x^2+2y^2-3z^2=-100\)

\(=>2\left(\dfrac{3y}{4}\right)^2+2y^2-3\left(\dfrac{5y}{4}\right)^2=-100\)

\(=>y=\pm8\)

* với y=8 \(=>\left\{{}\begin{matrix}x=\dfrac{3.8}{4}=6\\z=\dfrac{5.8}{4}=10\end{matrix}\right.\)

* với y=-8=>\(\left\{{}\begin{matrix}x=-6\\z=-10\end{matrix}\right.\)