Bài 2: 

\(\Leftrightarrow n+1\in\left\{1;2;4\right\}\)

hay \(n\in\left\{0;1;3\right\}\)

a) \(\left(2+x\right)\left(-1+2y\right)=5\)
Từ đó ta có bảng sau:

 b) \(3xy-y+3x=5\)
\(\Rightarrow y\left(3x-1\right)+\left(3x-1\right)=4\)
\(\Rightarrow\left(y+1\right)\left(3x-1\right)=4\)
Từ đó ta có bảng sau:

đây nhé ; k nha Tìm x,y là số nguyên thỏa mãn: 3xy - 5 = x2 + 2y

3xy+2y=2-x

=>3xy+2y+x=2

=>\(y\left(3x+2\right)+x+\dfrac{2}{3}=2+\dfrac{2}{3}=\dfrac{8}{3}\)

=>\(3y\left(x+\dfrac{2}{3}\right)+\left(x+\dfrac{2}{3}\right)=\dfrac{8}{3}\)

=>\(\left(x+\dfrac{2}{3}\right)\left(3y+1\right)=\dfrac{8}{3}\)

=>\(\left(3x+2\right)\left(3y+1\right)=8\)

=>\(\left(3x+2;3y+1\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-\dfrac{1}{3};\dfrac{7}{3}\right);\left(2;0\right);\left(-1;-3\right);\left(-\dfrac{10}{3};-\dfrac{2}{3}\right);\left(0;1\right);\left(\dfrac{2}{3};\dfrac{1}{3}\right);\left(-\dfrac{4}{3};-\dfrac{5}{3}\right);\left(-2;-1\right)\right\}\)

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