Ta có :
+) \(2x=3y\Rightarrow\)\(\dfrac{x}{3}=\dfrac{y}{2}\) \(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\)\(\left(1\right)\)
+) \(5x=7z\Rightarrow\dfrac{x}{7}=\dfrac{z}{5}\Rightarrow\dfrac{x}{21}=\dfrac{z}{15}\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\) Ta có :
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}\Rightarrow\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{15}=\dfrac{3x+7y+5z}{63+98+15}=\dfrac{30}{40}=\dfrac{3}{4}\)
Khi đó:
\(\dfrac{x}{21}=\dfrac{3}{4}\Rightarrow x=\dfrac{63}{4}\)
\(\dfrac{y}{14}=\dfrac{3}{4}\Rightarrow y=\dfrac{21}{2}\)
\(\dfrac{z}{14}=\dfrac{3}{4}\Rightarrow z=\dfrac{45}{24}\)
Vậy : \(\left[{}\begin{matrix}x=\dfrac{63}{4}\\y=\dfrac{21}{2}\\z=\dfrac{45}{24}\end{matrix}\right.\) là giá trị cần tìm
Ta có:\(2x=3y;3y=7z\Rightarrow2x=3y=7z\Rightarrow\dfrac{2x}{42}=\dfrac{3y}{42}=\dfrac{7z}{42}\)
\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{6}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{30}\)
Theo tính chất DTSBN ta có:
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{6}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{30}=\dfrac{3x-7y+5z}{63-98+30}=\dfrac{-30}{-5}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.21=126\\y=6.14=84\\z=6.6=36\end{matrix}\right.\)
Vậy...
