Question

_{\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\) và 2x+3y-z =124}

Answer 1

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)

nên \(\dfrac{x}{15}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)

nên \(\dfrac{y}{20}=\dfrac{z}{28}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)

mà 2x+3y-z=124

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{124}{62}=2\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{15}=2\\\dfrac{y}{20}=2\\\dfrac{z}{28}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)

Answer 2

\(\dfrac{x}{3}=\dfrac{x}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{125}{62}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=28.2=56\end{matrix}\right.\)