Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

=>x(y+2)-3(y+2)=-5

=>(x-3)(y+2)=-5

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}

\(\dfrac{1}{3}< \dfrac{x}{y}< \dfrac{1}{2}\Rightarrow\dfrac{4}{12}< \dfrac{x}{y}< \dfrac{6}{12}\Rightarrow\dfrac{x}{y}=\dfrac{5}{12}\Rightarrow\dfrac{x}{5}=\dfrac{y}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{12}=\dfrac{2x}{10}=\dfrac{3y}{36}=\dfrac{2x+3y}{10+36}=\dfrac{19}{46}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{95}{46}\\y=\dfrac{114}{23}\end{matrix}\right.\)

Mà \(x,y\in Z\)

Vậy ko có x,y nguyên thỏa mãn đề

khó phết

=>x(2y+1)-3y-1,5=2,5

=>(y+0,5)(2x-3)=2,5

=>(2y+1)(2x-3)=5

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

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(1;-3\right);\left(-1;-1\right)\right\}\)

\(2xy+x-3y=4\)

\(\Leftrightarrow4xy+2x-6y=8\)

\(\Leftrightarrow4xy+2x-6y-3=5\)

\(\Leftrightarrow2x\left(2y+1\right)-3\left(2y+1\right)=5\)

\(\Leftrightarrow\left(2x-3\right)\left(2y+1\right)=5\)

Vậy pt có các cặp nghiệm nguyên \(\left(x;y\right)=\left(-1;-1\right);\left(1;-3\right);\left(2;2\right);\left(4;0\right)\)

\(xy+2x-3y=1\\ \Rightarrow x\left(y+2\right)-3y-6=1-6\\ \Rightarrow x\left(y+2\right)-3\left(y+2\right)=-5\\ \Rightarrow\left(x-3\right)\left(y+2\right)=-5\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(-2;-1\right);\left(2;3\right);\left(4;-7\right);\left(8;-3\right)\right\}\)