Question

Cho x, y nguyên dương thỏa mãn \(2x^2y-4x^2+y-5=0\) Tính
x+y

Answer 1

\(2x^2y-4x^2+y-5=0\Leftrightarrow2x^2\left(y-2\right)+\left(y-2\right)=3\)

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

Do \(x,y\in Z^+,2x^2+1\ge1>0\Rightarrow y-2\in Z^+\) và \(y-2\inƯ\left(3\right)\)

\(\Rightarrow\left[{}\begin{matrix}y-2=3\\y-2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=5\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=3\\x=1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)\in\left\{5;4\right\}\)