x+y+xy=0

<=> xy+x+y=0

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

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

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

đến đây dễ rồi,bn tự giải tp nhé

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

\(\Leftrightarrow1+x+y.\left(1+x\right)=0+1\)

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

mà \(x,y\in Z\Rightarrow1+x;y+1\in Z\)

 Ta có: \(1=1.1=\left(-1\right).\left(-1\right)\)

\(\Rightarrow\hept{\begin{cases}1+x=1\\1+y=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}1+x=1\\1+y=1\end{cases}}\\\hept{\begin{cases}1+y=-1\\1+x=-1\end{cases}}\end{cases}}\) hoặc \(\hept{\begin{cases}1+x=-1\\1+y=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}\)               hoặc \(\hept{\begin{cases}x=-2\\y=-2\end{cases}}\)

Vậy \(\left(x,y\right)\in\left\{\left(0;0\right);\left(-2;-2\right)\right\}\)

Bn ơi ...Sao ko cho xy có khác nhau hay giống nhau ạ

nếu giống nhau thì xy =0

Ta có: ( x - 2) x ( y + 3) = -13 = (-13) x 1 = (-1) x 13

* Nếu x - 2 = -13 => x = (-13) + 2 = -11

         y + 3 = 1 => y = 1-3 = -2 

* Nếu x-2 = -1 => x = (-1) + 2 = 1

         y + 3 = 13 => y = 13 - 3 = 10

Vậy có 2 cặp x;y x;y(-11;-2)

                          x;y(1;10)

a: =>x^2006-x=0

=>x(x^2005-1)=0

=>x=0 hoặc x=1

b: =>5^x*26=650

=>5^x=25

=>x=2

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

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

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

=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(0;-2\right);\left(-2;0\right)\right\}\)

 Từ x+y/13=x-y/3 => 3(x +y) = 13(x-y) => y = 5x/8 
Tu x-y/3=xy/200 => 200(x-y) = 3xy => 200(x - 5x/8) = 3x.5x/8 => x^2 - 40x => x(x-40) = 0 => x = 0 hoac x = 40. 
Voi x = 0 ta co y = 0 
Voi x = 40 ta co y = 25

chắc vậy đó

a. Ta thay : |x| > hoac = 0 => 3|x| > hoac = 0

Tuong tu 2|y| > hoac = 0 

Ma 3|x|+2|y| = 0

=> 3|x| = 0 => x = 0

=> 2|y| = 0 => y=0

Vay: x=y=0