Tìm các cặp số nguyên (x;y) sao cho  (2x + 1)(y – 3) = -6

Câu hỏi:

Tìm các cặp số nguyên (x;y) sao cho  (2x + 1)(y – 3) = -6

Ta có: (2x + 1)(y – 3) = -6

=>  (2x + 1)(y – 3) = -1.6 = 1.(-6)=2.(-3)=(-2).3

* (2x + 1)(y – 3) = -1.6

\( \Rightarrow \left\{ \begin{array}{l}
2x + 1 =  – 1\\
y – 3 = 6
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x =  – 1\\
y = 9
\end{array} \right.\)

Hoặc \( \Rightarrow \left\{ \begin{array}{l}
2x + 1 = 6\\
y – 3 =  – 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{5}{2}\\
y = 2
\end{array} \right.\)

* (2x + 1)(y – 3) = 1. (-6)

\( \Rightarrow \left\{ \begin{array}{l}
2x + 1 = 1\\
y – 3 =  – 6
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 0\\
y =  – 3
\end{array} \right.\)

Hoặc \( \Rightarrow \left\{ \begin{array}{l}
2x + 1 =  – 6\\
y – 3 =  – 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x =  – \frac{7}{2}\\
y = 4
\end{array} \right.\)

* (2x + 1)(y – 3) = 2. (-3)

\( \Rightarrow \left\{ \begin{array}{l}
2x + 1 = 2\\
y – 3 =  – 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{1}{2}\\
y = 0
\end{array} \right.\)

Hoặc \( \Rightarrow \left\{ \begin{array}{l}
2x + 1 = 6\\
y – 3 =  – 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{5}{2}\\
y = 2
\end{array} \right.\)

* (2x + 1)(y – 3) = (-2).3

\( \Rightarrow \left\{ \begin{array}{l}
2x + 1 =  – 2\\
y – 3 = 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{{ – 3}}{2}\\
y = 6
\end{array} \right.\)

Hoặc \( \Rightarrow \left\{ \begin{array}{l}
2x + 1 = 3\\
y – 3 =  – 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\)

Vậy tìm được 4 cặp số nguyên (x; y) là (-1; 9); (0; -3); (-2; 5); (1; 1)