Lời giải:

Với $x,y$ nguyên thì $x-2, 2y+3$ nguyên. Mà $(x-2)(2y+3)=26$, $2y+3$ lẻ với mọi $y$ nguyên nên ta xét các TH sau:

TH1: $2y+3=1, x-2=26\Rightarrow y=-1; x=28$

TH2: $2y+3=-1, x-2=-26\Rightarrow y=-2; x=-24$

TH3: $2y+3=13, x-2=2\Rightarrow y=5, x=4$

TH4: $2y+3=-13, x-2=-2\Rightarrow y=-8; x=0$

\(\left(x-2\right)\left(2y+3\right)=26\)

\(\Leftrightarrow x-2\inƯ\left(26\right);2y+3\inƯ\left(26\right)\) \(=\left\{\pm1;\pm2;\pm13;\pm26\right\}\)

Vì \(\left(2y+3\right):2\) dư 1 nên \(2y+3=\pm1;\pm13\) 

Lại có \(x-2;2y+3\inℕ\) nên \(2y+3=13\)

 Khi \(2y+3=13\) ; \(x-2=2\)

      \(\Rightarrow y=5;x=4\)

Vậy cặp ( x;y) thỏa mãn đề bài là : \(x=4;y=5\)

a: =>(x-1)(y+2)=4

=>\(\left(x-1;y+2\right)\in\left\{\left(1;4\right);\left(4;1\right);\left(2;2\right);\left(-1;-4\right);\left(-4;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(5;-1\right);\left(3;0\right);\left(0;-6\right);\left(-3;-3\right)\right\}\)

b: =>x(2y+1)=36

=>\(\left(x;2y+1\right)\in\left\{\left(36;1\right);\left(12;3\right);\left(4;9\right);\left(-36;-1\right);\left(-12;-3\right);\left(-4;-9\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(36;0\right);\left(12;1\right);\left(4;4\right);\left(-36;-1\right);\left(-12;-2\right);\left(-4;-5\right)\right\}\)

(x-3)(2y+1)=7 

\(\Rightarrow\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=3\end{cases}}}\)(thỏa)

\(\hept{\begin{cases}x-3=-1\\2y+1=-7\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-4\end{cases}}}\)(thỏa)

Vậy nếu x=4 thì y=3

x=2 thì y=-4

\(\left(x-3\right)\left(2y+1\right)=7\)

\(\Rightarrow x-3;2y+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

Ta có bảng sau :

Vậy các cặp số nguyên (x,y) là : (4,3) ; (2,-4) ; (10,0) ; (-4,-1)

sao cau nay nhieu nguoi noi the

x=3;y=1

x = 3 ; y = 1

x=3

y=1