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

\(\Leftrightarrow x\left(y+4\right)+\left(y+4\right)=7\)

\(\Leftrightarrow\left(x+1\right)\left(y+4\right)=7\)

Vì x ; y nguyên nên x + 1 nguyên , y + 4 nguyên

Ta có bảng

Vậy ,.............

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

\(\Rightarrow x\left(y+4\right)+\left(y+4\right)=3+4\)

\(\Rightarrow\left(x+1\right)\left(y+4\right)=7\)

\(\Rightarrow\left(x+1\right);\left(y+4\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có các trường hợp sau 

\(TH1:\hept{\begin{cases}x+1=1\\y+4=7\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=3\end{cases}}}\)            \(TH2:\hept{\begin{cases}x+1=-1\\y+4=-7\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-11\end{cases}}}\)

\(TH3:\hept{\begin{cases}x+1=7\\y+4=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=-3\end{cases}}}\)      \(TH4:\hept{\begin{cases}x+1=-7\\y+4=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-8\\y=-5\end{cases}}}\)

Vậy\(\left(x;y\right)\in\left\{\left(0;3\right);\left(-2;-11\right);\left(6;-3\right);\left(-8;-5\right)\right\}\)

\(a)\)\(xy-x-y=1\)

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

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

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

\(\Rightarrow\)\(\left(x-1\right);\left(y-1\right)\inƯ\left(2\right)\)

Lập bảng : 

Vậy \(\left(x,y\right)\in\left\{\left(2;3\right),\left(3;2\right),\left(0;-1\right),\left(-1;0\right)\right\}\)
Chúc bạn học tốt ~ 

\(b)\)\(xy-2x-2y=1\)

\(\Leftrightarrow\)\(\left(xy-2x\right)-\left(2y-4\right)=5\)

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

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

\(\Rightarrow\)\(\left(x-2\right);\left(y-2\right)\inƯ\left(5\right)\)

Lập bảng : 

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

Chúc bạn học tốt ~ 

4:

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

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

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

Ta có : xy - 4x - 3y = 5

=> xy - 4x - 3y + 12 = 5 + 12

=> x(y - 4) - 3(y - 4) = 17

=> (x - 3)(y - 4) = 17

Vì x;y \(\inℤ\Rightarrow x-3;y-4\inℤ\)

Khi đó ta có 17 = 1.17 = (-1).(-17)

Lập bảng xét các trường hợp 

Vậy các cặp (x;y) thỏa mãn là (4;21) ; (20;5) ; (2;-13) ; (-14;3)

a;\(xy+3x-y=8\)

\(\Rightarrow x\left(y+3\right)-\left(y+3\right)=8-3\)

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

\(\Rightarrow\left(x-1\right);\left(y+3\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Xét bảng 

Vậy..............................

b,\(2xy-4x+y=8\)

\(\Rightarrow x\left(2y-4\right)+y=8\)

\(\Rightarrow2x\left(2y-4\right)+\left(2y-4\right)=8-4\)

\(\Rightarrow\left(2x+1\right)\left(2y-4\right)=4\)

\(\Rightarrow\left(2x+1\right);\left(2y-4\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Xét bảng 

Vậy.....................................