x +  3y = xy + 3

=> -xy + x = -3y + 3

=> x[-y + 1] = 3[-y + 1]

=> x = 3

Vậy với mọi y và x = 3 thì ta đc pt đúng

x+3y=xy+3

=> 3y-3=xy-x

=> 3(y-1)=x(y-1)

=> 3=x

=> x=3

NX : 3+3y=3y+3

=> với x=3 thì y là các giá trị nào cũng thỏa mãn .

VD: 3+3.2=3.2+3=9 ;...

x + 3y = xy + 3
=> -xy + x = -3y + 3
=> x[ -y + 1] = 3[- y + 1]
=> x = 3
Vậy y và x bằng 3 thì ta được phương trình đúng.

\(x+xy+y=1\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(Vậy...\)

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

\(a,\Leftrightarrow y\left(x+1\right)-3\left(x+1\right)=5\\ \Leftrightarrow\left(x+1\right)\left(y-3\right)=5=5.1=\left(-5\right)\left(-1\right)\\ TH_1:\left\{{}\begin{matrix}x+1=1\\y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=8\end{matrix}\right.\\ TH_2:\left\{{}\begin{matrix}x+1=5\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\\ TH_3:\left\{{}\begin{matrix}x+1=-5\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=2\end{matrix}\right.\\ TH_4:\left\{{}\begin{matrix}x+1=-1\\y-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(0;8\right);\left(4;4\right);\left(-6;2\right);\left(-2;-2\right)\right\}\)

\(b,\Leftrightarrow6\left(n-1\right)+11⋮n-1\\ \Leftrightarrow n-1\in\left\{-11;-1;1;11\right\}\\ \Leftrightarrow n\in\left\{-10;0;2;12\right\}\)

 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

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 !!!!!

 Ta có :

\(x^2+y^2+xy=3\)

\(\Rightarrow\left(x+y\right)^2-xy=3\)

\(\Rightarrow \left(x+y\right)^2=3+xy\)

hay \(S^2=3+xy\le3+\frac{\left(x+y\right)^2}{4}=3+\frac{S^2}{4}\)

\(\Rightarrow S^2\le3+\frac{S^2}{4}\)

\(\Rightarrow S^2\le4\)

\(\Rightarrow-2\le S\le2\)

GTLN của S = 2