\(3x=5y=10z\Leftrightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{10}}\)
\(\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{2y}{\dfrac{2}{5}}=\dfrac{z}{\dfrac{1}{10}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{2y}{\dfrac{2}{5}}=\dfrac{z}{\dfrac{1}{10}}=\dfrac{x-2y-z}{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{1}{10}}=\dfrac{15}{-\dfrac{1}{6}}=-90\)
\(\Rightarrow\left\{{}\begin{matrix}x=-90.\dfrac{1}{3}=-30\\y=-90.\dfrac{1}{5}=-18\\z=-90.\dfrac{1}{10}=-9\end{matrix}\right.\)
3x = 5x = 10z và x-2y-z=15
Ta có
\(\dfrac{3x}{30}=\dfrac{5y}{30}=\dfrac{10z}{30}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{3}\)
Đặt \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{3}=k\Rightarrow x=10k;y=6k;z=3k\)
Thay k vào x-2y-z=15
\(\Rightarrow10k-2\cdot6k-3k\)
\(\Rightarrow k\left(10-12-3\right)=15\)
\(\Rightarrow k\cdot\left(-5\right)=15\)
\(\Rightarrow k=-3\)
Từ đó suy ra
x=10k=-30
y=6k=-18
z=3k=-9
Vây x=-30;y=-18;z=-9
