Question

\(2x=3y,5y=7z\) và \(3x-7y=5z=30\)

Answer 1

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:

\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.21\\y=2.14\\z=2.10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\)

Answer 2

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 ............................