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

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

\(\Leftrightarrow\frac{xy}{2y}-\frac{2}{2y}=\frac{1}{3}\)

\(\Leftrightarrow\frac{xy-2}{2y}=\frac{1}{3}\)

\(\Leftrightarrow3\left(xy-2\right)=2y\)

\(\Leftrightarrow3xy-6=2y\)

\(\Leftrightarrow3xy-2y=6\)

\(\Leftrightarrow y\left(3x-2\right)=6=1.6=2.3=3.2=6.1=\left(-1\right).\left(-6\right)=\left(-3\right).\left(-2\right)\)

                                                                                               \(=\left(-2\right).\left(-3\right)=\left(-6\right).\left(-1\right)\)

Tự xét bảng giá trị nhé

đúg đề hết chưa bn

Ta có: x-2xy+y-3=0

=>-2xy+x+y=3

=>-2.(-2xy+x+y)=-2.3

=>4xy-2x-2y=-6

=>4xy-2x-2y+1=-6+1

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

=>(2y-1)(2x-1)=-5=1.(-5)=-5.1=(-1).5=5.(-1)

Ta có bảng sau:

Vậy (x;y) E {(1;-2);(-2;1);(3;0);(0;3)}

<=> \(x=\frac{3-y}{1+2y}\)

Để x, y\(\in\)Z\(1+2y\in\text{Ư}_5\)hì \(\frac{3-y}{1+2y}\in Z\)

<=>\(3-y⋮1+2y\)

<=>\(-1-2y-5⋮1+2y\)

<=>\(1+2y\in\text{Ư}\)5

<=>\(1+2y\in\text{{}\text{ }1;-1;5;-5\)

<=>\(2y\in\text{{}0;-2;4;-6\)<=>\(y\in\text{{}0;-1;2;-3\)=>x=...

sao x lại bằng 3-x/1+2y vậy bạn

Ta thấy \(x-y+2xy=3\Rightarrow x\left(1+2y\right)=3+y\Rightarrow x=\frac{3+y}{1+2y}\)

\(x\in Z\Rightarrow2x\in Z\). Ta có \(2x=\frac{6+2y}{1+2y}=\frac{1+2y+5}{1+2y}=1+\frac{5}{1+2y}\)

\(\Rightarrow1+2y\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(y\in\left\{-3;-1;0;2\right\}\)

Với y = -3, \(x=0\)

Với y = -1, \(x=-2\)

Với y = 0, x = 3

Với y = 2, x = 1.