Đặt k = \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Khi đó : k3 = \(\frac{x}{2}\frac{y}{3}\frac{z}{5}=\frac{xyz}{2.3.5}=\frac{810}{30}=27\)
=> k = 3
Nên : \(\frac{x}{2}=3\Rightarrow x=6\)
\(\frac{y}{3}=3\Rightarrow y=9\)
\(\frac{z}{5}=3\Rightarrow z=15\)
Vậy x = 6 , y = 9 , z = 15
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)= k => \(x=2k\); \(y=3k\); \(z=5k\)
=> \(x.y.z=2k.3k.5k=30k^3=180\)=> \(k^3=\frac{180}{30}=6\)=> \(k=\sqrt[3]{6}\)
=> \(x=2\sqrt[3]{6}\); \(y=3\sqrt[3]{6}\); \(z=5\sqrt[3]{6}\)
gọi x/2=y/3=z/5=k
vì x/2=k =>x=2.k
vì y/3=k =>y=3.k
vì z/5=k =>z=5.k
thay x.y.z vào xyz=810
2k.3k.5k=810
2.3.5.k^3=810
30.k^3=810
k^3=810:30=27
=> k=3
x=2k=>x=2.3=6
y=3k=>y=3.3=9
z=5k=>z=5.3=15
Ta có \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\Rightarrow xyz=30k^3=810\Rightarrow k^3=27\Rightarrow k=3\Rightarrow x=6;y=9;z=15\)
Ta có : \(\frac{x}{2}=\frac{y}{3}\Leftrightarrow3x=2y\Leftrightarrow x=\frac{2y}{3}\) (1)
\(\frac{y}{3}=\frac{z}{5}\Leftrightarrow5y=3z\Leftrightarrow z=\frac{5y}{3}\) (2)
Thay (1) ; (2) vào biểu thức xyz = 810 ; ta được :
\(\frac{2y}{3}\cdot y\cdot\frac{5y}{3}=810\Leftrightarrow\frac{10y^3}{9}=810\Leftrightarrow10y^3=7290\Leftrightarrow y^3=729\Leftrightarrow y^3=9^3\Leftrightarrow y=9\)
- Với y = 9 => x = \(\frac{2.9}{3}=6\)
- Với y = 9 => z = \(\frac{5.9}{3}=15\)
Vậy x = 6 ; y = 9 và z = 15
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k,y=3k,z=5k\)
Thay \(x=2k;y=3k;z=5k\)vào \(xyz=810\), ta có:
\(2k.3k.5k=810\)
\(\Leftrightarrow k^3.30=810\)
\(\Leftrightarrow k^3=27=3^3\)
\(\Leftrightarrow k=3\)
\(x=2k=2.3=6\)
\(y=3k=3.3=9\)
\(z=5k=5.3=15\)
