Lời giải:
$xy+12=x+y$

$\Rightarrow xy-x-y=-12$

$\Rightarrow x(y-1)-y=-12$

$\Rightarrow x(y-1)-(y-1)=-11$

$\Rightarrow (y-1)(x-1)=-11$

Do $x,y$ nguyên nên $x-1,y-1$ cũng nguyên. Ta có bảng:

4:

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

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

=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

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

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

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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\(xy\) - \(x\) + \(y\) = 1

(\(xy\) + \(y\)) - \(x\) - 1 = 0

\(y\)(\(x\) + 1) - ( \(x\) + 1) = 0

(\(x\) + 1)( \(y\) - 1) = 0

\(\left[{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\)

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