Ta có: \(\dfrac{x-y}{1}=\dfrac{x+y}{7}=\dfrac{\left(x-y\right)+\left(x+y\right)}{1+7}=\dfrac{2x}{8}=\dfrac{x}{4}\)
\(\dfrac{x}{y}=\dfrac{xy}{24}\Leftrightarrow\dfrac{6x}{24}=\dfrac{xy}{24}\)
\(\Rightarrow6x=xy\)
\(\Rightarrow y=6\)
\(\dfrac{x-6}{1}=\dfrac{x+6}{7}\)
\(\Leftrightarrow7.\left(x-6\right)=x+6\)
\(\Leftrightarrow7x-42=x+6\)
\(\Leftrightarrow7x-x=6+42\)
\(\Leftrightarrow6x=48\)
\(\Rightarrow x=8\)
Vậy \(x=8;y=6.\)
Đặt : \(\dfrac{x-y}{1}=\dfrac{x+y}{7}=\dfrac{xy}{24}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=k\\x+y=7k\\xy=24k\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)
Từ (1) => x = y + k
Thay x = y + k vào (2)
=> 2y = 6k => y = 3k
Có : x = y + k ; y = 3k
=> xy = (y + k).3k = 24k
<=> y + k = 8
Mà y = 3k
=> 4k = 8
=> k = 2
\(\Rightarrow\left\{{}\begin{matrix}x-y=2\\x+y=14\\xy=48\end{matrix}\right.\)(tới đây lập luận , thế qua thế lại là ra)
