d) 2xy+x+y=4

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

ta có bảng sau

thử lại ta có x=0 ,y=4

h cho minh nha

Nếu thế thì x=4, y=0 cx đc

a) \(x-2xy-3y+1=0\)

\(\Leftrightarrow x-y\left(2x-3\right)+1=0\)

\(\Leftrightarrow2x-2y\left(2x-3\right)+2=0\)

\(\Leftrightarrow2x-2y\left(2x-3\right)-3+5=0\)

\(\Leftrightarrow-2y\left(2x-3\right)+\left(2x-3\right)=-5\)

\(\Leftrightarrow\left(2x-3\right)\left(1-2y\right)=-5\)

tự tìm nốt

b) \(x-2xy+y=0\)

\(\Leftrightarrow x\left(1-2y\right)+y=0\)

\(\Leftrightarrow-2x\left(1-2y\right)-2y=0\)

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

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

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

Tìm nốt

x.(1-2y)+2y=-4

x(1-2y)  +  1.(1-2y)=  -4+1

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

ta có bag kq

tik đúg nha

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

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

\(x\).(2y - 1) = 2 + y

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

vì \(x\) ∈ Z ⇔ (2 + y) ⋮ (2y - 1)

2(2+ y) ⋮ (2y - 1)

(2y + 4) ⋮ (2y - 1)

(2y -1 + 5) ⋮ (2y - 1)

5 ⋮ (2y - 1)

(2y - 1) ∈ Ư(5) = {-5; -1; 1; 5}

Lập bảng ta có:

"Theo bảng trên ta có:

(\(x;y\)) = (0; -2); (-2; 0); (3; 1); (1; 3)

Vậy: (x;y) = (0; -2); (-2; 0); (3; 1); (1; 3)

2xy-x-y=2

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

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

=>\(\left(2x-1\right)\left(y-\frac12\right)=\frac52\)

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

=>(2x-1;2y-1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(2x;2y)∈{(2;6);(6;2);(0;-4);(-4;0)}

=>(x;y)∈{(1;3);(3;1);(0;-2);(-2;0)}

ai trả lời đc mình k cho

\(\text{x ‐ 2 xy + y = 0}\)

\(\text{=> x‐(2xy‐y)=0}\)

\(\text{=> x‐ y(2x‐1)=0}\)

\(\text{=> (2x‐2y)(2x‐1)=0}\)

\(\text{=> ( 2x‐1) ‐2y(2x‐1)=‐1}\)

\(\text{=> (2x‐1)(1‐2y)=‐1}\)

\(\text{=> ( 2x‐1 ; 1‐2y ) = ( ‐1 ;1 ﴿ ; ﴾1;‐1 )}\)

\(\text{=> (x;y)=( 0 ; 0 ) ; ( 1;1)}\)