Theo bài ra ta có :
\(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{3}=\dfrac{7y}{14}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) \(\left(1\right)\)
\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{2y}{14}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\Rightarrow\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x+5z-7y}{63+50-98}=\dfrac{30}{15}=2\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x}{63}=2\Rightarrow\dfrac{x}{21}=2\Rightarrow x=42\\\dfrac{7y}{98}=2\Rightarrow\dfrac{y}{14}=2\Rightarrow y=28\\\dfrac{5z}{50}=2\Rightarrow\dfrac{z}{10}=2\Rightarrow z=20\end{matrix}\right.\\ \)
\(\text{Vậy }x=42\\ y=28\\ z=20\)
Ta có:
\(2x=3y\Rightarrow10x=15y\)
\(5y=7z\Rightarrow15y=21z\)
\(\Rightarrow10x=15y=21z\Rightarrow\dfrac{10x}{210}=\dfrac{15y}{210}=\dfrac{21z}{210}\)
\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x+5z-7y}{3.21+5.14-7.10}\)
\(=\dfrac{30}{63+70-70}=\dfrac{30}{63}=\dfrac{10}{21}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{21}.21=10\\y=\dfrac{10}{21}.14=\dfrac{20}{3}\\z=\dfrac{10}{21}.10=\dfrac{100}{21}\end{matrix}\right.\)
Chúc bạn học tốt!!!
