Giải :
Ta có : \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\\\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Đặt \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=k\Rightarrow\left\{{}\begin{matrix}x=10k\\y=15k\\z=21k\end{matrix}\right.\)
Thay x = 10k ; y = 15k và z = 21k vào x + y + z = 92 ta có :
10k + 15k + 21k = 92
⇔ k ( 10+15+21) = 92
\(\Leftrightarrow k\cdot46=92\\ \Leftrightarrow k=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\cdot2=20\\y=15\cdot2=30\\z=21\cdot2=42\end{matrix}\right.\\ Vậyx=20;y=30vàz=42\)
ta có :\(\dfrac{x}{2}=\dfrac{y}{3}=>\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)
lại có :\(\dfrac{y}{5}=\dfrac{z}{7}=>\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)
từ (1)và (2) => \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
=>\(\left\{{}\begin{matrix}x=20\\y=30\\z=42\end{matrix}\right.\)
chúc bạn học tốt
