1/ Ta có: 6x = 4y = 3z
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}\)
Đặt \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=4k\\y=3k\\z=6k\end{matrix}\right.\)
Ta có: 4k + 3k + 6k = 18
13k = 18
k = \(\dfrac{18}{13}\) (1)
Thay (1) vào, ta có
\(\left\{{}\begin{matrix}x=4.\dfrac{18}{13}\\y=3.\dfrac{18}{13}\\z=6.\dfrac{18}{13}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{72}{13}\\y=\dfrac{54}{13}\\z=\dfrac{108}{13}\end{matrix}\right.\)
2/Ta có: 3x = 2y = z
\(\Rightarrow\dfrac{x}{2}=y=\dfrac{z}{3}\)
Đặt \(\dfrac{x}{2}=y=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=k\\z=3k\end{matrix}\right.\)
Ta có: 2k + k + 3k = 99
6k = 99
k = \(\dfrac{33}{2}\) (1)
Thay (1) vào , ta có:
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{33}{2}\\y=\dfrac{33}{2}\\z=3.\dfrac{33}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=33\\y=\dfrac{33}{2}\\z=\dfrac{99}{2}\end{matrix}\right.\)
Tương tự câu a,b, bn tự làm tiếp câu c nhé
