Question

[Đỗ Việt Hùng]

Tìm x,y,z biết: \(\frac{x-1}{2}=\frac{y+3}{4}\frac{z-5}{6}\) và 5z - 3x - 4y =50

Answer 1

[fan FA]

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)và 5z-3x-4y=50

ta có:

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=>\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-16-6}\)\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)

\(=\frac{50-34}{8}=\frac{16}{8}=2\)

Vậy x=5;y=5;z=17

Answer 2

[Vũ Quang Vinh]

Theo đầu bài ta có:
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)
\(\Rightarrow\frac{3\left(x-1\right)}{3\cdot2}=\frac{4\left(y+3\right)}{4\cdot4}=\frac{5\left(z-5\right)}{5\cdot6}\)
\(\Rightarrow\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)
\(=\frac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)
\(=\frac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}x=2\cdot2+1=5\\y=2\cdot4-3=5\\z=2\cdot6+5=17\end{cases}}\)