Ta có: x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4

xy=xy24⇔6x24=xy24xy=xy24⇔6x24=xy24

⇒6x=xy⇒6x=xy

⇒y=6⇒y=6

x−61=x+67x−61=x+67

⇔7.(x−6)=x+6⇔7.(x−6)=x+6

⇔7x−42=x+6⇔7x−42=x+6

⇔7x−x=6+42⇔7x−x=6+42

⇔6x=48⇔6x=48

⇒x=8⇒x=8

Vậy x=8;y=6

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

bạn cho mình hỏi x,y có là số tự nhiên không 

có bạn nhé

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)

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

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...

\(P=\dfrac{x^3+y^3}{x^3y^3}=\dfrac{\left(x+y\right)\left(x^2+y^2-xy\right)}{x^3y^3}=\dfrac{x^2y^2\left(x+y\right)}{x^3y^3}=\dfrac{x+y}{xy}=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{x^2+y^2-xy}=\dfrac{4\left(x^2+y^2-xy\right)-3\left(x^2+y^2-2xy\right)}{x^2+y^2-xy}\)

\(=4-\dfrac{3\left(x-y\right)^2}{x^2+y^2-xy}\le4\)

\(P_{max}=4\) khi \(x=y=\dfrac{1}{2}\)