Ta có:
\(2x=3y\)
\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) (1)
\(5y=7z\)
\(\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) (2)
Từ (1) và (2) suy ra: \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.21\\y=2.14\\z=2.10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\)
Theo bài ra ta có:
\(+)2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\)
\(+)5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\)
\(\Leftrightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{21}=2\Rightarrow x=42\\\dfrac{y}{14}=2\Rightarrow y=28\\\dfrac{z}{10}=2\Rightarrow z=20\end{matrix}\right.\)
Vậy ............................
