Tìm x, y biết 6x=4y = 3z và x+y+z = 18

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