\(Đặt\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k,\)\(\) ta có: \(x=2k;y=3k;z=5k\)
Vì \(x.y.z=-240\Rightarrow2k.3k.5k=-240\)
\(\Rightarrow30k^3=-240\Rightarrow k^3=-240:30=-8\)
\(\Rightarrow k^3=\left(-2\right)^3\Rightarrow k=-2\)
\(\)Ta có:
\(x=2k\Rightarrow x=-2.2=-4\)
\(y=3k\Rightarrow y=-2.3=-6\)
\(z=5k\Rightarrow z=-2.5=-10\)
Vậy \(x=-4;y=-6;z=-10\)
Từ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x.y.y}{2.3.5}=\dfrac{-240}{30}\) = \(-8\)
=> \(\dfrac{x}{2}=\left(-8\right);\dfrac{y}{3}=\left(-8\right);\dfrac{z}{5}=\left(-8\right)\)
Với : \(\dfrac{x}{2}=\left(-8\right)\Rightarrow x=-16\)
Với:\(\dfrac{y}{3}=\left(-8\right)\Rightarrow y=-24\)
Với:\(\dfrac{z}{5}=\left(-8\right)\Rightarrow z=-40\)