Question

Tìm x: \(\dfrac{4}{x-3}=\dfrac{8}{y-6}=\dfrac{20}{z-15}\) và xyz= 640 Help me!!!

Answer 1

\(\dfrac{4}{x-3}=\dfrac{8}{y-6}=\dfrac{20}{z-15}\)

\(\Rightarrow\dfrac{x-3}{4}=\dfrac{y-6}{8}=\dfrac{z-15}{20}\)

\(\Rightarrow\dfrac{x}{4}-\dfrac{3}{4}=\dfrac{y}{8}-\dfrac{3}{4}=\dfrac{z}{20}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{8}=\dfrac{z}{20}\)

Đặt: \(\dfrac{x}{4}=\dfrac{y}{8}=\dfrac{z}{20}=k\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=8k\\z=20k\end{matrix}\right.\)

Thay vào đk đề bài: \(640k^3=640\Leftrightarrow k=1\)

Vậy \(\left\{{}\begin{matrix}x=4\\y=8\\z=20\end{matrix}\right.\)