mk nói đại nha
Áp dụng tính chất của dãy tỉ số bằng nhau ta có
x.y.z/3.5.6=720/90=8
=>
x/3=8=> x= 3.8=24
y/5=8=> y= 5.8=40
z/6=8 => z= 6.8=48
vậy: ...
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}\text{ và }x.y.z=720\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=6k\end{matrix}\right.\)
\(\text{Mà }x.y.z=720\)
\(\Rightarrow3k.5k.6k=720\)
\(\Rightarrow90.k^3=720\)
\(\Rightarrow k^3=720:90\)
\(\Rightarrow k^3=8\)
\(\Rightarrow k^3=2^3\)
\(\Rightarrow k=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k=3.2=6\\y=5k=5.2=10\\z=6k=6.2=12\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=6\\y=10\\z=12\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}\) và \(x.y.z=720\)
đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=6k\end{matrix}\right.\)
\(\Rightarrow3k.5k.6k=720\)
\(\Rightarrow90\).k\(^3\) \(=720\)
\(\Rightarrow k^3=720:90=8=(\pm2)^3\)
\(\Rightarrow k=\pm2\)
nếu \(k=2\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=12\end{matrix}\right.\)
nếu \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-6\\y=-10\\z=-12\end{matrix}\right.\)
vậy nếu \(x=6\Rightarrow y=10\Rightarrow z=12\)
nếu \(x=-6\Rightarrow y=-10\Rightarrow z=-12\)
