Question

Answer 1

a, Ta có : \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{-2x}{6}=\dfrac{y}{-5}=\dfrac{3z}{-12}=\dfrac{3z-2x}{6-12}=-6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-6\\\dfrac{y}{-5}=-6\\\dfrac{z}{-4}=-6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=30\\z=24\end{matrix}\right.\)

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b, Ta có : \(\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{15}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{15}=\dfrac{x+y+z}{4+6+15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=30\end{matrix}\right.\)

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