Question

tìm x;y;z biết x/3 = y/4 = z/5 và z - x + y = 48

Answer 1

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{z-x+y}{5-3+4}=\frac{48}{6}\text{ }=8\)

\(\Rightarrow\frac{x}{3}=8\Rightarrow x=24\)

\(\frac{y}{4}=8\Rightarrow y=32\)

\(\frac{z}{5}=8\Rightarrow z=40\)

Answer 2

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

Ta có \(z-x+y=48\Rightarrow5k-3k+4k=48\Rightarrow6k=48\Rightarrow k=8\)

\(\Rightarrow\hept{\begin{cases}x=3k=3.8=24\\y=4k=4.8=32\\z=5k=5.8=40\end{cases}}\)

Vậy x = 24; y = 32; z = 40