`a,2xy+6x+y=2`
`⇒ x(2y-6)+y=2`
`⇒ x(2y-6)+y.2-6=2.2-6`
`⇒ x(2y-6)+(2y-6)=-2`
`⇒ (2y-6)(x+1)=-2`
Ta có bảng sau :
\begin{array}{|l|r|} \hline x+1 &1&-1&2&-2\\ \hline 2y-6 &-2&2&-1&1\\ \hline x&0&-2\quad(L)&1&-3\quad(L) \\ \hline y&2&///\quad&2,5\quad(L)&///\quad \\ \hline\end{array}
Vậy `(x;y)=(0;2)`
`b,2xy+2x+3y=2`
`⇒ 2x(y+1)+3y+3=2+3`
`⇒ 2x(y+1)+3(y+1)=5`
`⇒ (y+1)( 2x+3)=5`
Ta có bảng sau :
\begin{array}{|l|r|} \hline 2x+3 &1&5&-1&-5\\ \hline y+1 &5&1&-5&-1\\ \hline x&1&-1\quad(L)&-2\quad(L)&-8\quad(L) \\ \hline y&0&///\quad&///\quad&///\quad \\ \hline\end{array}
Vậy `(x;y)=(1;0)`
Xin hay nhất !
