Đáp án:
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Có : `2x = 3y`
`↔ x/3 = y/2`
`↔ x/21 = y/14` `(1)`
Có : `7z = 5y`
`↔z/5 = y/7`
`↔ z/10 = y/14` `(2)`
Từ `(1), (2)`
`↔ x/21 = y/14 = z/10`
`↔ (3x)/63 = (7y)/98 = (5z)/50`
Áp dụng tính chất dãy tỉ số bằng nhau có :
`(3x)/63 = (7y)/98 = (5z)/50 = (3x - 7y + 5z)/(63 - 98 + 50) = 30/15 = 2`
`↔` \(\left\{ \begin{array}{l}\dfrac{3x}{63}=2\\ \dfrac{7y}{98}=2\\ \dfrac{5z}{50}=2\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}3x=63×2\\7y=98×2\\5z=50×2\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}3x=126\\7y=196\\5z=100\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=126÷3\\y=196÷7\\z=100÷5\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=42\\y=28\\z=20\end{array} \right.\)
Vậy `(x;y;z) = (42;28;20)`
