برای مجموعه های\(A = \{ a - 2,6,2b + 1,c\} \) و\(B = \{ \sqrt d ,5, - 1\} \) ، فرض کنید \(A \times B = B \times A\)باشد.در چند حالت مقدار\(a + b + c = 9\) است ؟

می دانیم برای اینکه \(A \times B = B \times A\)  باشد لازم است A=B. یعنی :

\(\begin{array}{l}\{ a - 2,6,2b + 1,c\} = \{ \sqrt d ,5, - 1\} \\\end{array}\)

\(\sqrt d = 6 \Rightarrow d = 36\)

اولا A باید دارای 3 عضو باشد،پس :

\( \Rightarrow a - 2 = 6 \Rightarrow a = 8\)حالت اول

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{1)\left\{ {\begin{array}{*{20}{l}}{2b + 1 = 5 \Rightarrow b = 2}\\{c = - 1}\end{array}} \right. \Rightarrow a + b + c = 9 \to true}\\{2)\left\{ {\begin{array}{*{20}{l}}{2b + 1 = - 1 \Rightarrow b = - 1}\\{c = 5}\end{array}} \right. \Rightarrow a + b + c = 12 \to untrue}\end{array}} \right.\)

\( \Rightarrow 2b + 1 = 6 \Rightarrow b = \frac{5}{2}\)حالت دوم

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{1)\left\{ {\begin{array}{*{20}{l}}{a - 2 = 5 \Rightarrow a = 7}\\{c = - 1}\end{array}} \right. \Rightarrow a + b + c = 8.5 \to untrue}\\{2)\left\{ {\begin{array}{*{20}{l}}{a - 2 = - 1 \Rightarrow a = 1}\\{c = 5}\end{array}} \right. \Rightarrow a + b + c = 8.5 \to untrue}\end{array}} \right.\)

\( \Rightarrow c = 6\)حالت سوم

\( \Rightarrow \left\{ \begin{array}{l}1)\left\{ \begin{array}{l}2b + 1 = 5 \Rightarrow b = 2\\a - 2 = - 1 \Rightarrow a = 1\end{array} \right. \Rightarrow a + b + c = 9 \to true\\2)\left\{ \begin{array}{l}a - 2 = 5 \Rightarrow a = 7\\2b + 1 = - 1 \Rightarrow b = - 1\end{array} \right. \Rightarrow a + b + c = 12 \to untrue\end{array} \right.\)

البته ممکن است حالات زیر نیز رخ دهد.

\(1 \Rightarrow \left\{ \begin{array}{l}a - 2 = c = 5 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{c = 5}\end{array}} \right.\\2b + 1 = - 1 \Rightarrow b = - 1\end{array} \right. \Rightarrow a + b + c = 11 \to untrue\)

\(2 \Rightarrow \left\{ \begin{array}{l}a - 2 = c = - 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{c = - 1}\end{array}} \right.\\2b + 1 = 5 \Rightarrow b = 2\end{array} \right. \Rightarrow a + b + c = 2 \to untru\)

\(3 \Rightarrow \left\{ \begin{array}{l}2b + 1 = c = 5 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{c = 1}\\{b = 2}\end{array}} \right.\\a - 2 = - 1 \Rightarrow a = 1\end{array} \right. \Rightarrow a + b + c = 8 \to untrue\)

\(4 \Rightarrow \left\{ \begin{array}{l}2b + 1 = c = - 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{c = - 1}\\{b = - 1}\end{array}} \right.\\a - 2 = 5 \Rightarrow a = 7\end{array} \right. \Rightarrow a + b + c = 5 \to untrue\)

\(5 \Rightarrow \left\{ \begin{array}{l}2b + 1 = a - 2 = 5 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{b = 2}\\{a = 7}\end{array}} \right.\\c = - 1\end{array} \right. \Rightarrow a + b + c = 8 \to untrue\)

\(6 \Rightarrow \left\{ \begin{array}{l}2b + 1 = a - 2 = - 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{b = - 1}\\{a = 1}\end{array}} \right.\\c = 5\end{array} \right. \Rightarrow a + b + c = 5 \to untrue\)