\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9};9z=7y\Rightarrow\dfrac{z}{7}=\dfrac{y}{9}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}\)
Áp dụng...
\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x}{45}=\dfrac{2y}{18}=\dfrac{4z}{28}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\\ \Rightarrow\left\{{}\begin{matrix}x=-150\\y=-90\\z=-70\end{matrix}\right.\)
\(3x=5y\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}\)
hay \(\dfrac{x}{15}=\dfrac{y}{9}\left(1\right)\)
7y=9z
nên \(\dfrac{y}{9}=\dfrac{z}{7}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{4}=\dfrac{3x-2y-4z}{45-18-16}=\dfrac{10}{11}\)
Do đó: \(x=\dfrac{150}{11};y=\dfrac{90}{11};z=\dfrac{40}{11}\)
\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}\)
\(9z=7y\Rightarrow\dfrac{y}{9}=\dfrac{z}{7}\)
Áp dung t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x}{45}=\dfrac{2y}{18}=\dfrac{4z}{28}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).15=-150\\y=\left(-10\right).9=-90\\z=\left(-10\right).7=-70\end{matrix}\right.\)
