\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{10}\)
\(=\dfrac{5x+y-2z}{50+6-10}=\dfrac{8}{46}=\dfrac{4}{43}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{43}.10=\dfrac{40}{43}\\y=\dfrac{4}{43}.6=\dfrac{24}{43}\\z=\dfrac{4}{43}.5=\dfrac{20}{43}\end{matrix}\right.\)
Ta có: \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{5}\Rightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{10}\)
Áp dụng tc dãy tỉ số bằng nhau:
\(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{10}=\dfrac{5x+y-2z}{50+6-10}=\dfrac{4}{23}\)
Do \(\left\{{}\begin{matrix}\dfrac{5x}{50}=\dfrac{4}{23}\\\dfrac{y}{6}=\dfrac{4}{23}\\\dfrac{2z}{10}=\dfrac{4}{23}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{40}{23}\\y=\dfrac{24}{23}\\z=\dfrac{20}{23}\end{matrix}\right.\).
Vậy ...
Ta có :
\(5x+y-2z=8\)
\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{5}\Leftrightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2x}{42}=\dfrac{5x+y-2z}{50+6-10}=\dfrac{8}{46}=\dfrac{4}{23}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5x}{50}=\dfrac{4}{23}\Leftrightarrow x=\dfrac{40}{23}\\\dfrac{y}{6}=\dfrac{4}{23}\Leftrightarrow x=\dfrac{24}{23}\\\dfrac{z}{5}=\dfrac{4}{23}\Leftrightarrow z=\dfrac{20}{23}\end{matrix}\right.\)
Vậy ....
ta có : \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{5}\) áp dụng tính chất dãy tỉ số bằng nhau
ta có : \(\dfrac{5x+y-2x}{50+6-10}=\dfrac{8}{46}=\dfrac{4}{23}\)
\(\Rightarrow\) \(\dfrac{x}{10}=\dfrac{4}{23}\) \(\Leftrightarrow x=\dfrac{4}{23}.10=\dfrac{40}{23}\)
\(\Rightarrow\dfrac{y}{6}=\dfrac{4}{23}\Leftrightarrow y=\dfrac{4}{23}.6=\dfrac{24}{23}\)
\(\Rightarrow\dfrac{z}{5}=\dfrac{4}{23}\Leftrightarrow z=\dfrac{4}{23}.5=\dfrac{20}{23}\)
