Theo đề bài, ta có:
\(\dfrac{3x}{4}=\dfrac{y}{2}=\dfrac{3z}{5}\) và x - z = 15
\(\Rightarrow\dfrac{3x}{4}=\dfrac{y}{2}\Rightarrow6x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}\) (1)
\(\Rightarrow\dfrac{y}{2}=\dfrac{3z}{5}\Rightarrow5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\) (2)
(1)(2) \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}=\dfrac{x-z}{4-5}=-\dfrac{15}{1}=-15\)
\(\Rightarrow x=-60;y=-90;z=-75\)
\(\Rightarrow x+y+z=-225\)
\(\frac{3.x}{4}=\frac{y}{2}=\frac{3.z}{5}=3.\frac{x}{4}=\frac{y}{2}=3.\frac{z}{5}\)và x-z=15
\(3.\frac{x}{4}=\frac{y}{2}=3.\frac{z}{5}=3.\frac{x}{4}-3.\frac{z}{5}\)
\(=3.\left(\frac{x-z}{4-5}\right)=3.\left(\frac{15}{-1}\right)=3.-15=-45\)
\(\Rightarrow\frac{3.x}{4}=-45\Rightarrow x=-60\)
\(\Rightarrow\frac{y}{2}=-45\Rightarrow y=-90\)
\(\Rightarrow\frac{3.z}{5}=-45\Rightarrow z=-75\)
\(\Rightarrow\)x+y+z=-60+-90+-75=-225
