Ta có:
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\) (1)
\(\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{25}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{25}\)
Áp dụng tính chất dãy số bằng nhau ta có:
\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{25}=\dfrac{2x-y+4z}{2\cdot20-10+4\cdot25}=\dfrac{270}{130}=\dfrac{27}{13}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{27}{13}\Rightarrow x=\dfrac{540}{13}\)
\(\Rightarrow\dfrac{y}{10}=\dfrac{27}{13}\Rightarrow y=\dfrac{270}{13}\)
\(\Rightarrow\dfrac{z}{25}=\dfrac{27}{13}=\dfrac{675}{13}\)
Có: \(\dfrac{x}{10}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{10}=\dfrac{2y}{10}\left(1\right)\)
\(\dfrac{y}{2}=\dfrac{z}{3}\Leftrightarrow\dfrac{2y}{10}=\dfrac{2z}{15}\left(2\right)\)
Từ (1) và (2) => \(\dfrac{x}{10}=\dfrac{2y}{10}=\dfrac{2z}{15}\)=> \(\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{2z}{15}\)
Áp dung tính chất của dãy tỉ số bằng nhau và 2x - y + 4z = 270, ta có:
\(\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{2z}{15}=\dfrac{2x}{20}=\dfrac{4z}{30}=\dfrac{2x-y+4z}{20-5+30}=\dfrac{270}{45}=6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=6\\\dfrac{y}{5}=6\\\dfrac{2z}{15}=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=30\\z=45\end{matrix}\right.\)
Vậy...
Ta có:
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\left(1\right)\)
\(\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\Leftrightarrow\dfrac{2x}{40}=\dfrac{y}{10}=\dfrac{4z}{60}=\dfrac{2x-y+4z}{40-10+60}=\dfrac{270}{90}=3\)
\(\Rightarrow\dfrac{x}{20}=3\Rightarrow x=60\)
\(\Rightarrow\dfrac{y}{10}=3\Rightarrow y=30\)
\(\Rightarrow\dfrac{z}{15}=3\Rightarrow z=45\)
