áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{-3}\)=\(\dfrac{y}{2}\)=\(\dfrac{z}{5}\)=\(\dfrac{xyz}{\left(-3\right).2.5}\)=\(\dfrac{240}{-30}\)=-8
=>\(\dfrac{x}{-3}\)=-8;\(\dfrac{y}{2}\)=-8
\(\dfrac{x}{-3}\)=-8=>x=-8.-3=24
\(\dfrac{y}{2}\)=-8=>y=-8.2=>-16
vậy x=24,y=-16
Đặt \(\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3k\\y=2k\\z=5k\end{matrix}\right.\)
\(\Rightarrow xyz=-30k^3\)
Mà \(xyz=240\Rightarrow-30k^3=240\Rightarrow k^3=-8\Rightarrow k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3k=-3.\left(-2\right)=6\\y=2k=2.\left(-2\right)=-4\end{matrix}\right.\)
Vậy \(x=6;y=-4\)
\(\dfrac{x}{-3}=\dfrac{y}{2}\Rightarrow y=-\dfrac{2}{3}x\)
\(\dfrac{x}{-3}=\dfrac{z}{5}\Rightarrow z=-\dfrac{5}{3}x\)
Thế \(y=-\dfrac{2}{3}x;z=-\dfrac{5}{3}x\) vào biểu thức \(xyz=240\), ta có:
\(x.\left(-\dfrac{2}{3}x\right).\left(-\dfrac{5}{3}\right)x=240\)
\(x^3.\dfrac{10}{9}=240\)
\(x^3=240:\dfrac{10}{9}=216\)
\(x=6\)
\(\Rightarrow y=-\dfrac{2}{3}.6=-4\)
Vậy \(x=6;y=-4\)
