x+3y = xy+3
<=> x+3y-xy-3 = 0
(x-3)+(3y-xy)=0
(x-3)+y.(3-x)=0
(x-3)-y.(x-3)=0
(x-3).(1-y)=0
=>x-3=0 ; 1-y=0
=>x=3 ; y=1
\(x+3y=xy+3\)
\(\Leftrightarrow x+3y-xy-3=0\)
\(\Leftrightarrow\left(x-xy\right)+\left(3y-3\right)=0\)
\(\Leftrightarrow x.\left(1-y\right)+3.\left(y-1\right)=0\)
\(\Leftrightarrow x.\left(1-y\right)-3.\left(1-y\right)=0\)
\(\Leftrightarrow\left(1-y\right).\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-y=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-0\\x=0+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)
Vậy cặp số nguyên x ; y thỏa mãn là \(\left(3;1\right).\)
Chúc bạn học tốt!
