Gọi cd,cr lần lượt là a,b(m;a,b>0)
Ta có \(a:b=5:3\Rightarrow\dfrac{a}{5}=\dfrac{b}{3}\)
Đặt \(\dfrac{a}{5}=\dfrac{b}{3}=k\Rightarrow a=5k;b=3k\)
Mà \(ab=960\Rightarrow15k^2=960\Rightarrow k^2=64\)
\(\Rightarrow k=8\left(k>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a=40\\b=24\end{matrix}\right.\)
Vậy ...
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)
Ta có: xy=960
\(\Leftrightarrow k=8\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=40\\y=3k=24\end{matrix}\right.\)