Đáp án đúng:
Giải chi tiết:Yêu cầu Giải Câu toán tương đương \({a^2} - 3\) chia hết cho \(ab + 3\)\(\begin{array}{l} \Rightarrow b\left( {{a^2} - 3} \right) \vdots \left( {ab + 3} \right) \Rightarrow \left[ {a\left( {ab + 3} \right) - 3\left( {a + b} \right)} \right] \vdots \left( {ab + 3} \right)\\ \Rightarrow 3\left( {a + b} \right) = k\left( {ab + 3} \right),k \in \mathbb{N}*\end{array}\)Nếu \(k = 1 \Rightarrow 3\left( {a + b} \right) = ab + 3 \Rightarrow \left( {a - 3} \right)\left( {b - 3} \right) = 6\)\(Do\,\,a,b \in \mathbb{N}* \Rightarrow \left\{ \begin{array}{l}a - 3 \ge  - 2\\b - 3 \ge  - 2\end{array} \right.\)\(TH1:\left[ \begin{array}{l}\left\{ \begin{array}{l}a - 3 = 6\\b - 3 = 1\end{array} \right.\\\left\{ \begin{array}{l}a - 3 = 1\\b - 3 = 6\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 9\\b = 4\end{array} \right.(tm)\\\left\{ \begin{array}{l}a = 4\\b = 9\end{array} \right.(ktm)\end{array} \right.\);\(TH2:\left[ \begin{array}{l}\left\{ \begin{array}{l}a - 3 = 3\\b - 3 = 2\end{array} \right.\\\left\{ \begin{array}{l}a - 3 = 2\\b - 3 = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 6\\b = 5\end{array} \right.(tm)\\\left\{ \begin{array}{l}a = 5\\b = 6\end{array} \right.(ktm)\end{array} \right.\)Nếu \(k = 2 \Rightarrow 3\left( {a + b} \right) = 2\left( {ab + 3} \right) \Rightarrow \left( {2a - 3} \right)\left( {2b - 3} \right) =  - 3\)\(Do\,a,b \in \mathbb{N}* \Rightarrow \left[ \begin{array}{l}2a - 3 \ge  - 1\\2b - 3 \ge  - 1\end{array} \right.\)Ta có: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}2a - 3 = 3\\2b - 3 =  - 1\end{array} \right.\\\left\{ \begin{array}{l}2a - 3 =  - 1\\2b - 3 = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 3\\b = 1\end{array} \right.\\\left\{ \begin{array}{l}a = 1\\b = 3\end{array} \right.\end{array} \right.\). Thử lại thì \(\left( {a;b} \right) = \left( {3;1} \right)\) thỏa mãn.Nếu \(k \ge 3 \Rightarrow 3\left( {a + b} \right) = k\left( {ab + 3} \right) \ge 3\left( {ab + 3} \right) \Leftrightarrow \left( {a - 1} \right)\left( {b - 1} \right) + 2 < 0\)(vô lý vì \(a,b \in \mathbb{N}*)\)Vậy các cặp số \(\left( {a,b} \right)\) thỏa mãn là \(\left( {3;1} \right);\left( {6;5} \right);\left( {9;4} \right)\)