Đặt x/12 = y/9 = z/5 = k ta có:
x = 12k
y = 9k
z = 5k
=> x.y.z = 12k.9k.5k
=> k^3.540=20
=> k^3 = 1/27
=> k^3= (1/3)^3
=> k = 1/3
x/12=1/3 => x=4
y/9= 1/3 => y=3
z/5=1/3 =. z=5/3
Gọi \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow x=12k;y=9k;z=5k\)
\(\Rightarrow xyz=12k.9k.5k=540k^3=20\)
\(\Rightarrow k^3=\frac{20}{540}=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow k=\frac{1}{3}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{3}\Rightarrow x=\frac{1}{3}.12=4\)
\(\frac{y}{9}=\frac{1}{3}\Rightarrow y=\frac{1}{3}.9=3\)
\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=\frac{1}{3}.5=\frac{5}{3}\)
Vậy \(x=4;y=3;z=\frac{5}{3}\)
Đặt x/12 = y/9 = z/5 = k
Ta có : x = 12k ; y = 9k ; z=5k
Ta có : x.y.z = 20 => 12k.9k.5k = 20
=> k3 = 20:5:9:12
=> k3 = (1/3)3
=> k = 1/3
=> x/12 = 1/3 => x = 1/3.12 = 4
=> y/9 = 1/3 => y = 1/3.9 = 3
=> z/5 = 1/3 => z = 1/3.5 = 5/3
Vậy ( x;y;z ) = ( 4;3;5/3 )
Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(=>\hept{\begin{cases}\frac{x}{12}=k=>x=12k\\\frac{y}{9}=k=>y=9k\\\frac{z}{5}=k=>z=5k\end{cases}}\)
Thay vào ta có :
\(x.y.z=20\)
\(=>12k.9k.5k=20\)
\(=>540k^3=20\)
\(=>k^3=\frac{1}{27}\)
\(=>k^3=\left(\frac{1}{3}\right)^3\)
\(=>k=\frac{1}{3}\)
Từ đó \(=>\hept{\begin{cases}x=12k=\frac{12}{3}=4\\y=9k=\frac{9}{3}=3\\z=5k=\frac{5}{3}\end{cases}}\)
Vậy ...
Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k=>\hept{\begin{cases}x=12k\\y=9k\\z=5k\end{cases}}\)
Thay vào :\(xyz=20\)
\(=>540k^3=20\)
\(=>k^3=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(=>k=\frac{1}{3}\)
Nên :\(\hept{\begin{cases}x=\frac{1}{3}.12=4\\y=\frac{1}{3}.9=3\\z=\frac{1}{3}.5=\frac{5}{3}\end{cases}}\)
Vậy ...