Question

Tìm x , y , z biết :

\(\begin{cases} \dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\\ x+y+z=49 \end{cases}\)

Answer 1

\(\dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

\(\Rightarrow\dfrac{2}{5}x=\dfrac{3}{4}y=\dfrac{4}{5}z\)

\(\Rightarrow\dfrac{2}{5}x.\dfrac{1}{12}=\dfrac{3}{4}y.\dfrac{1}{12}=\dfrac{4}{5}z.\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}\)

Đặt \(\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=30k\\y=16k\\z=15k\end{matrix}\right.\). Ta có:

\(x+y+z=49\)

\(\Rightarrow30k+16k+15k=49\)

\(\Rightarrow61k=49\)

\(\Rightarrow k=\dfrac{49}{61}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)