ko can tl dau, mk biet lam rui may bn ak . Dung tl nha

Ta có : \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{2x}{8}=\frac{x}{4}\)

\(\frac{x}{4}=\frac{xy}{24}\Leftrightarrow\frac{6x}{24}=\frac{xy}{24}\) => 6x = xy => y = 6

\(\frac{x-6}{1}=\frac{x+6}{7}\)

<=> 7(x - 6) = x + 6

<=> 7x - 42 = x + 6

<=> 7x - x = 6 + 42

<=> 6x = 48

=> x = 8

Vậy x = 8 ; y = 6

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)

xy=48

48

xy=48

48

=> \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{xy}{24}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{xy}{24}=\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{\left(x-y\right)-\left(x+y\right)}{1-7}\)=> \(\frac{xy}{24}=\frac{x}{4}=\frac{y}{3}\)

\(\frac{xy}{24}=\frac{x}{4}\)=>\(\frac{x}{4}.\frac{y}{6}=\frac{x}{4}\)=>  \(\frac{y}{6}=\frac{x}{4}:\frac{x}{4}=1\) ( do x khác 0) => y = 6

\(\frac{xy}{24}=\frac{y}{3}\Rightarrow\frac{x}{8}.\frac{y}{3}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{3}:\frac{y}{3}=1\) ( do y khác 0) => x = 8

Vậy...