a) \(\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\ \dfrac{y}{2}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{15}\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{15}=\dfrac{x+3y-2z}{8+6.3-2.15}=\dfrac{36}{-4}=-9\)
\(\dfrac{x}{8}=-9\Rightarrow x=-72\\ \dfrac{y}{6}=-9\Rightarrow y=-54\\ \dfrac{z}{15}=-9\Rightarrow z=-135\)
b) \(9x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\\ 2x=z\Rightarrow\dfrac{x}{1}=\dfrac{z}{2}\Rightarrow\dfrac{x}{5}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{10}\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{10}=\dfrac{x+3y-2z}{5+9.3-2.10}=\dfrac{36}{12}=3\)
\(\dfrac{x}{5}=3\Rightarrow x=15\\ \dfrac{y}{9}=3\Rightarrow y=27\\ \dfrac{z}{10}=3\Rightarrow z=30\)
\(2x=3y=4z\Rightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3.\dfrac{1}{3}-2.\dfrac{1}{4}}=\dfrac{36}{\dfrac{1}{3}}=108\)
\(\dfrac{x}{\dfrac{1}{2}}=108\Rightarrow x=54\\ \dfrac{y}{\dfrac{1}{3}}=108\Rightarrow y=36\\ \dfrac{z}{\dfrac{1}{4}}=108\Rightarrow z=27\)
