Question

[amano ichigo]

tìm x,y,z biết: x/2= y/3= z/5 với x.y.z=-240

Answer 1

[Hoài Thanh Dương]

\(Đặt\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k,\)\(\) 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 \(x=-4;y=-6;z=-10\)

Answer 2

[Miinhhoa]

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:\(\dfrac{z}{5}=\left(-8\right)\Rightarrow z=-40\)