\(y+2⋮x;x+2⋮y\Rightarrow\left(x+2\right)\left(y+2\right)⋮xy\Rightarrow xy+2x+2y+4⋮xy\Rightarrow2x+2y+4⋮xy\)

\(\Rightarrow2\left(x+y+2\right)⋮xy\Rightarrow2⋮xy\Rightarrow xy\inƯ\left(2\right)=1;2\)

\(xy=1\Rightarrow x=1,y=1\Rightarrow y+2=1+2=3⋮x=1\Rightarrow y+2⋮x\)

                                             \(x+2=1+2=3⋮y=1\Rightarrow x+2⋮y\)

\(\Rightarrow x=1,y=1\left(tm\right)\)

\(xy=2\Rightarrow x=1,y=2;x=2,y=1\Rightarrow x+2=1+2=3\)ko chia hết cho \(y=2\Rightarrow x+2\)ko chia hết cho y

\(\Rightarrow x=1,y=2\left(ktm\right)\Rightarrow x=2,y=1\left(ktm\right)\)

vậy x=1,y=1 

ko chắc lắm

\(\Rightarrow2^y\left(2^{x-y}+1\right)=72\)

Vì \(2^{x-y}+1\) lẻ nên \(2^y\left(2^{x-y}+1\right)=72=2^3\cdot9\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\\2^{x-3}+1=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=3\\2^{x-3}=8=2^3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(6;3\right)\)

Ta có: \(1024=2^{10}\)

Từ đề bài ta suy ra được \(x>y\)

Gỉa sử \(x=k+y\left(k>0\right)\), ta có:

\(2^{y+k}-2^y=2^y.2^k-2^y=2^y\left(2^k-1\right)\)

\(\Rightarrow2^y\left(2^k-1\right)=2^{10}\)

\(\Rightarrow2^k-1=2^{10-y}\)

Vì \(2^k-1\) là số lẻ nên \(2^k-1⋮̸2\left(k\ne0\right)\) 

Mà \(2^{10-y}⋮2\)(sai) \(\Rightarrow k=0;y=10\)

\(\Rightarrow x=10+0=10\)

Vậy \(x=y=10\)

hai cái x,y đều bằng 10 vậy trừ cho nhau = 0 rồi bn ơi

1024=2¹⁰\(\Rightarrow\)2^y.(2^m-1)=2¹⁰.1\(\Rightarrow\hept{\begin{cases}x=11\\y=10\end{cases}}\)

 Vậy x=11;y=10

 Ai ngang qua đừng quên để lại 1 L_I_K_E!!!!

\(1024=2^{10}\)\(\Rightarrow2^y\left(2^m-1\right)=2^{10}.1\Rightarrow\hept{\begin{cases}y=10\\x=11\end{cases}}\)

=> x>y

x-y =m

mk bít lm nè