Xét \(2x^2+3x+2=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{16}>0\forall x\in R\)

=> \(x^3< y^3\left(1\right)\) (1)

Giả sử : \(y^3< \left(x+2\right)^3\)

\(\Leftrightarrow x^3+2x^2+3x+2< x^3+6x^2+12x+8\)

\(\Leftrightarrow-4x^2-9x-6< 0\)

\(\Leftrightarrow4x^2+9x+6>0\)

\(\Leftrightarrow4\left(x+\dfrac{9}{8}\right)^2+\dfrac{15}{64}>0\)

=> Giả sử đúng .

=> \(y^3< \left(x+2\right)^3\left(2\right)\)

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

\(\Leftrightarrow x^3+2x^2+3x+2=x^3+3x^2+3x+1\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

.) Khi \(x=1\Rightarrow y=2\).

.) Khi \(x=-1\Rightarrow y=0\)

Vậy nghiệm của pt ( x;y ) = {( 1;2 ) ; ( -1;0 )}

\(2\left(xy-3\right)=x\)

\(\Leftrightarrow2xy-6=x\)

\(\Leftrightarrow2xy-x=0+6\)

\(\Leftrightarrow x\left(2y-1\right)=6\)

\(\Rightarrow x\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\Rightarrow y\in\left\{....\right\}\)

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

\(\Leftrightarrow x^2y+2y=x^3+3x-5\)

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

\(\Leftrightarrow y=\frac{x^3+3x-5}{x^2+2}=\frac{x^3+2x+x-5}{x^2+2}\)

\(=\frac{x\left(x^2+2\right)+\left(x-5\right)}{x^2+2}=\frac{x\left(x^2+2\right)}{x^2+2}+\frac{x-5}{x^2+2}\)

\(=x+\frac{x-5}{x^2+2}\)

Mà \(x,y\in Z\)

\(\Rightarrow\frac{x-5}{x^2+2}\in Z\)

\(\Rightarrow x-5⋮x^2+2\)

\(\Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\)

\(\Rightarrow x^2-25⋮x^2+2\)

\(\Rightarrow x^2+2-27⋮x^2+2\)

\(\Rightarrow27⋮x^2+2\)

\(\Rightarrow\left(x^2+2\right)\inƯ\left(27\right)\)

Mà \(Ư\left(27\right)=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)

Nhưng \(x^2+2\ge2\forall x\)

\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)

Lập bảng giá trị :

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{\pm1;\pm5\right\}\)        \(\left(2\right)\)

Thay \(\left(2\right)\)vào   \(\left(1\right)\)ta có :

+) Với \(x=-1\Rightarrow y=-3\)    ( thõa mãn )

+) Với \(x=1\Rightarrow y=-\frac{1}{3}\)   ( loại )

+) Với \(x=-5\Rightarrow y=-\frac{145}{27}\)   ( loại )

+) Với \(x=5\Rightarrow y=5\)  ( thõa mãn )

Vậy các số nguyên \(\left(x,y\right)\)cần tìm là : \(\left(-1;-3\right)\) ;       \(\left(5;5\right)\)