\(8x^2-3xy-5y=25\)

\(\Leftrightarrow8x^2-25=3xy+5y\Leftrightarrow8x^2-25=y\left(3x+5\right)\)

\(\Leftrightarrow y=\frac{8x^2-25}{3x+5}\)\(\Rightarrow9y=\frac{72x^2-225}{3x+5}=24x-40-\frac{25}{3x+5}\)

\(\Rightarrow3x+5\inƯ\left(25\right)=\pm1;\pm5;\pm25\)

Đến đây bạn tự suy ra x rồi thay vào biểu thức trên để suy ra y là ok.

\(2\left(x+y\right)+1=3xy\)

=>\(2x+2y-3xy=1\)

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

=>\(-x\left(3y-2\right)+2y-\dfrac{4}{3}=-\dfrac{1}{3}\)

=>\(-3x\left(y-\dfrac{2}{3}\right)+2\left(y-\dfrac{2}{3}\right)=-\dfrac{1}{3}\)

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

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

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

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

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

mà x,y nguyên

nên (x,y)=(1;1)

Ta có:  

x² + 2y² + 3xy + 3x + 5y = 15

<=> x² + 2y² + 3xy + 3x + 5y + 2 = 17

<=> (x² + xy + 2x) + (2xy + 2y² + 4y) + (x + y + 2) = 17

<=> (x + y + 2)(x + 2y + 1) = 17

=> (x + y + 2, x + 2y + 1) = (1,17; 17,1; - 1,-17; -17,-1)

Giải ra là tìm được x,y nhé

VeryVery good.Thanks. I will give 1  for you.Love

\(x^2+2y^2+3xy-x-y+3=0\)

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

cần nx k mk gửi