\(pt\Leftrightarrow\left(x-1\right)y=x^2+2\)
\(+\text{Nếu }x-1=0\Leftrightarrow x=1\text{ thì }pt\text{ trở thành: }0=x^2+2\text{ (vô nghiệm)}\)
\(+\text{Xét }x-1\ne0\Leftrightarrow x\ne1\)
\(pt\Leftrightarrow y=\frac{x^2+2}{x-1}=\frac{x\left(x-1\right)+x-1+3}{x-1}=x+1+\frac{3}{x-1}\)
\(y\text{ nguyên nên }\frac{3}{x-1}\text{ nguyên}\Rightarrow x-1\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x\in\left\{-2;0;2;4\right\}\)
Vậy các nghiệm nguyên của pt là
\(\left(x;y\right)=\left(-2;-2\right);\left(0;-2\right);\left(2;6\right);\left(4;6\right)\)
x^2 - xy + y + 2 = 0
=> -xy + y +x^2 +2 = 0
=> -y[x-1] + x^2+2 = 0
=> y.[x-1] = x^2+2
=> y = [x^2+2] / [x-1 ] ; x
v
