\(x+xy+y=1\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(Vậy...\)

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)

<=> xy+x+y=0

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

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

Lập bảng là ra

lập hộ mik cái bảng

bạn thùy linh lập hộ cái bảng vs

a) x= 1

   y= 2

hihihi

CHTT ko có mà bn!!!

mà sao bn toàn CHTT thế

a)x= 1

   y= 2

 tick mk nhavõ kiều oanh

x(y+2)+y = 1

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

(y+2)(x+1) = 3

ta co bang

\(\Rightarrow2x-4xy+2y=0\\ \Rightarrow2x\left(1-2y\right)+2y-1=-1\\ \Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\\ \Rightarrow\left(2x-1\right)\left(2y-1\right)=1=1.1=\left(-1\right)\left(-1\right)\)

Với \(\left\{{}\begin{matrix}2x-1=1\\2y-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\rightarrow\left(1;1\right)\)

Với \(\left\{{}\begin{matrix}2x-1=-1\\2y-1=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\rightarrow\left(0;0\right)\)

Vậy các cặp \(\left(x;y\right)\) cần tìm là \(\left(1;1\right);\left(0;0\right)\)

|x - 7| + |y| = 2

Vì \(\hept{\begin{cases}\left|x-7\right|\ge0\\\left|y\right|\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-7\right|=x-7\\\left|y\right|=y\end{cases}}\)

Nên ta xét bảng sau

Vậy (x;y) = (8;1) = (9;0) = (7;2)

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 !!!!!