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Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\) =>
- \(x=3k\)
- \(y=4k\)
- \(z=5k\)
=> \(3k.4k.5k=x.y.z\)
=> \(\left(3.4.5\right).k^3=480\)
=> \(60.k^3=480\)
=> \(k^3=\frac{480}{60}=8\)
=> \(k=2\)
Từ đó suy ra: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=2\)
=> \(x=3.2=6\)
=> \(y=4.2=8\)
=> \(z=5.2=10\)
Đặt k = \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Rightarrow x=3k,y=4k,z=5k\)
Từ x.y.z = 480, ta có:
3k.4k.5k = 480
\(\Rightarrow60k^3\) = 480
\(\Rightarrow k^3=8\)
\(\Rightarrow k=2\)
Với k = 2 \(\Rightarrow\) x = 6, y = 8, z = 10
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Lời giải :
a) Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=5k\\y=4k\\z=2k\end{cases}}\)
Ta có : \(xyz=40k^3=240\)
\(\Leftrightarrow k^3=6\)
\(\Leftrightarrow k=\sqrt[3]{6}\)
\(\Leftrightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=\sqrt[3]{6}\)
\(\Leftrightarrow\hept{\begin{cases}x=5\sqrt[3]{6}\\y=4\sqrt[3]{6}\\z=2\sqrt[3]{6}\end{cases}}\)
Vậy....
b) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{9}=\frac{y}{6}\)
Ta cũng có \(\frac{y}{3}=\frac{z}{2}\Leftrightarrow\frac{y}{6}=\frac{z}{4}\)
Khi đó : \(\frac{x}{9}=\frac{y}{6}=\frac{z}{4}=\frac{x-y+z}{9-6+4}=\frac{21}{7}=3\)
\(\Leftrightarrow\hept{\begin{cases}x=27\\y=18\\z=12\end{cases}}\)
Vậy...
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