\(x+xy+y=1\)

\(2x+2xy+2y=2\)

\(2x\left(1+y\right)+2y=2\)

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

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

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

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

\(\left(x+1\right)\left(y+1\right)=2\)

\(TH1:\left\{{}\begin{matrix}x+1=1\\y+1=2\end{matrix}\right.\)

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

\(TH2:\left\{{}\begin{matrix}x+1=2\\y+1=1\end{matrix}\right.\)

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

\(TH3:\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}x+1=-2\\y+1=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

\(Vậy...\)

x+xy+y=1⇔x(y+1)+y+1=2⇔(x+1)(y+1)=2

⇒(x+1;y+1)=(-1;-2),(-2;-1),(1;2),(2;1)

sau tự tính nhé :3

xy -x - y =2

x.( y-1) = 2+ y

\(x=\frac{2+y}{y-1}=\frac{y-1+3}{y-1}=\frac{y-1}{y-1}+\frac{3}{y-1}=1+\frac{3}{y-1}\)

Để x nguyên

\(\Rightarrow\frac{3}{y-1}\in z\Rightarrow3⋮y-1\Rightarrow y-1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)

nếu y- 1 = 3 => y =4 (TM)  => x = 1+ 3/4-1 = 1 + 1 =2 => x= 2 (TM)

y-1 =-3 => y =-2 (TM) => x = 1+ 3/-2-1 = 1+(-1) =0 => x =0 (TM)

y -1 = 1 => y=2 (TM) => x = 1+ 3/2-1 = 1+3 =4 => x =4 (TM)

y-1 =-1 => y=0 (TM) => x = 1+ 3/0-1 = 1+(-3) = -2 => x = -1 (TM)

KL: (x;y) =........................................

Chúc bn học tốt !!!!!

                xy=x+y

nên :        xy-(x+y)=0

               xy-x-y    =0

               x(y-1)-y  =0 suy ra x(y-1)-(y-1)=1

                (x-1)(y-1)=1 

ta có

x=0 , y=0

x=2 , y=2

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

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

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

mà \(x,y\)là số nguyên nên ta có bảng giá trị: 

xy = x - y

=> xy - x + y = 0

=> x(y - 1) + (y - 1) = 0 - 1

=> (x + 1)(y - 1) = -1

=> x + 1 và y - 1 thuộc Ư(-1) = {1;-1}

Ta có bảng:

Vậy các cặp (x;y) là (0;0) ; (-2;2)

\(xy=-\left(x+y\right)\)

\(\Leftrightarrow xy+x+y=0\)

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

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

.....

x².(y+1) + y = 30

x². (y+1) + (y+1) = 29

(y+1).(x²+1) = 29 = 1 . 29 = 29 . 1