\(2x=3y=5z\Leftrightarrow\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x-y+z}{\frac{1}{2}-\frac{1}{3}+\frac{1}{5}}=-\frac{33}{\frac{15-10+6}{30}}=-\frac{33}{\frac{11}{30}}=-90\)
\(x=\frac{1}{2}\cdot\left(-90\right)=-45;y=\frac{1}{3}\cdot\left(-90\right)=-30;z=\frac{1}{5}\cdot\left(-90\right)=-18\)
Ta có 2x\(=\)3y\(=\)5z BCNN(2;3;5) \(=\)30 Nên \(\frac{2x}{30}\)\(=\)\(\frac{3y}{30}\)\(=\)\(\frac{5z}{30}\) Do đó \(\frac{x}{15}\)\(=\)\(\frac{y}{10}\)=\(\frac{z}{6}\) Áp dụng tính chất dãy tỉ số băng nhau và thay x\(-\)y\(+\)z \(=\)-33 ta đc \(\frac{x}{15}\) \(=\)\(\frac{y}{10}\)\(=\)\(\frac{z}{6}\)\(\frac{x-y+z}{15-10+6}\)\(=\)\(\frac{-33}{11}\)\(=\)-3 Do đó x\(=\)(-3)\(\times\)15 \(=\)-45 y \(=\)(-3) \(\times\)10\(=\)-30 z\(=\)(-3)\(\times\)6 \(=\)-18
