Đặt \(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{12}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=9k\\z=12k\end{matrix}\right.\)
=> x.y.z = 5k.9k.12k <=> 540k3 = 20
k3 = 20:540
k3 = \(\dfrac{1}{27}\)
<=> \(\dfrac{1}{27}=\dfrac{1^3}{3^3}\) => k = \(\dfrac{1}{3}\)
\(\left\{{}\begin{matrix}x=5k=5.\dfrac{1}{3}=\dfrac{5}{3}\\y=9k=9.\dfrac{1}{3}=3\\z=12k=12.\dfrac{1}{3}=4\end{matrix}\right.\)
Vậy x=5/3 ; y=3 và z=4
Lời giải:
Ta có:
\(\frac{x}{5}=\frac{y}{9}=\frac{z}{12}\Rightarrow \left(\frac{x}{5}\right)^3=\left(\frac{y}{9}\right)^3=\left(\frac{z}{12}\right)^3=\frac{x}{5}.\frac{y}{9}.\frac{z}{12}=\frac{20}{540}=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow \left\{\begin{matrix} \frac{x}{5}=\frac{1}{3}\\ \frac{y}{9}=\frac{1}{3}\\ \frac{z}{12}=\frac{1}{3}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{5}{3}\\ y=3\\ z=4\end{matrix}\right.\)
Vậy \((x,y,z)=(\frac{5}{3};3;4)\)
Đặt x5=y9=z12=k⇒⎧⎪⎨⎪⎩x=5ky=9kz=12kx5=y9=z12=k⇒{x=5ky=9kz=12k
=> x.y.z = 5k.9k.12k <=> 540k3 = 20
k3 = 20:540
k3 = 127127
<=> 127=1333127=1333 => k = 1313
⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x=5k=5.13=53y=9k=9.13=3z=12k=12.13=4{x=5k=5.13=53y=9k=9.13=3z=12k=12.13=4
Vậy x=5/3 ; y=3 và z=4
