Đặt \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{6}=k\)
=> \(\hept{\begin{cases}x=5k\\y=-4k\\z=6k\end{cases}}\) (1)
Khi đó, ta cóL
\(\left(5k\right).\left(-4k\right).\left(6k\right)=15\)
=> \(-120k^3=15\)
=> \(k^3=-\frac{1}{8}\)
=> \(k=-\frac{1}{2}\)
Thay k = -1/2 vào (1), ta được:
x = 5 . (-1/2) = -2,5
y = -4.(-1/2) = 2
z = 6 . (-1/2) = -3
Vậy ...
b)Đặt \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{6}=k\)
\(\Rightarrow x=5k;y=-4k;z=6k\)
\(\Rightarrow xyz=5k.\left(-4k\right).6k=-120k^3\)
\(\Rightarrow15=-120k^3\)
\(\Rightarrow k^3=-\frac{1}{8}\Rightarrow k=-\frac{1}{2}\)
Từ \(\frac{x}{5}=-\frac{1}{2}\Rightarrow x=5\)
\(\frac{y}{-4}=-\frac{1}{2}\Rightarrow y=2\)
\(\frac{z}{6}=-\frac{1}{2}\Rightarrow z=-3\)
Vậy x = 5 ; y = -2 ; z = -3
Đặt \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{6}=k\Rightarrow x=5k;y=-4k;z=6k\)
\(\Rightarrow xyz=5k.\left(-4k\right).6k=k^3.\left(-120\right)=15\)
\(\Rightarrow k^3=\frac{15}{-120}=\frac{-1}{8}=\left(\frac{-1}{2}\right)^3\)
\(\Rightarrow k=-\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}.5=\frac{-5}{2}\\y=-\frac{1}{2}.\left(-4\right)=2\\z=-\frac{1}{2}.6=-3\end{cases}}\)
