Question

tìm cac số nguyên sao cho 3x³ +xy =3

Answer 1

\(\Leftrightarrow\) x(3x²+y)=3

\(\Rightarrow\) x;3x²+y\(\in\) Ư(3)=\(\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\3x+y=3\Leftrightarrow3+y=3\Rightarrow y=0\end{matrix}\right. \)

⇒\(\left\{{}\begin{matrix}x=-1\\3x+y=-3\Leftrightarrow3+y=-3\Rightarrow y=-6\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.;\left\{{}\begin{matrix}x=-1\\y=-6\end{matrix}\right.\)