Question

[Kirigaya Kazuto]

Tìm \(x;y\) ϵ N :

a) \(\left(2x+1\right)\left(y-3\right)=10\)

b) \(x-3=y\left(x+2\right)\)

Answer 1

[Nguyễn Trọng Thắng]

a) (2x+1)(y-3)=10

\(\Rightarrow\)\(\begin{cases}\left(2x+1\right)=10\\\left(y-3\right)=10\end{cases}\) \(^{_{ }\Rightarrow}\) \(\begin{cases}x=4,5\\y=7\end{cases}\)

Vậy x= 4,5 và y=7

Answer 2

[Phùng Tuệ Minh]

a) (2x+1)(y-3)=10=1.10=10.1=2.5=5.2

\(\Rightarrow\left[{}\begin{matrix}2x+1=1;y-3=10\\2x+1=10;y-3=1\\2x+1=2;y-3=5\\2x+1=5;y-3=2\end{matrix}\right.\)

Lại có 2x+1 là số lẻ \(\Rightarrow\left[{}\begin{matrix}2x+1=1;y-3=10\\2x+1=5;y-3=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=13\\x=2;y=5\end{matrix}\right.\)

Vậy: \(\left(x;y\right)=\left(0;13\right)\left(2;5\right)\)