tim cac so nguyen x;y sao cho:a (x+2)(y-1)=3 b (3-x)(xy+5)=-1 AI TRA LOI DUNG VA NHANH NHAT MINH TICK CHO

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1. Đáp án:

   \[\begin{array}{l}
   a,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x;y} \right) \in \left\{ {\left( { – 1;4} \right);\left( { – 3; – 2} \right);\left( {1;2} \right);\left( { – 5;0} \right)} \right\}\\
   b,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x;y} \right) \in \left\{ {\left( {2; – 3} \right);\left( {4; – 1} \right)} \right\}
   \end{array}\]

   Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   a,\\
   \left( {x + 2} \right)\left( {y – 1} \right) = 3\\
   x,y \in Z \Rightarrow \left( {x + 2} \right);\left( {y – 1} \right) \in Z\\
   3 = 1.3 = \left( { – 1} \right).\left( { – 3} \right)\\
    \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x + 2 = 1\\
   y – 1 = 3
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 2 =  – 1\\
   y – 1 =  – 3
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 2 = 3\\
   y – 1 = 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 2 =  – 3\\
   y – 1 =  – 1
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x =  – 1\\
   y = 4
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 3\\
   y =  – 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 1\\
   y = 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 5\\
   y = 0
   \end{array} \right.
   \end{array} \right.\\
    \Rightarrow \left( {x;y} \right) \in \left\{ {\left( { – 1;4} \right);\left( { – 3; – 2} \right);\left( {1;2} \right);\left( { – 5;0} \right)} \right\}\\
   b,\\
   \left( {3 – x} \right)\left( {xy + 5} \right) =  – 1\\
   x;y \in Z \Rightarrow \left( {3 – x} \right);\left( {xy + 5} \right) \in Z\\
    \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   3 – x = 1\\
   xy + 5 =  – 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   3 – x =  – 1\\
   xy + 5 = 1
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = 2\\
   xy =  – 6
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 4\\
   xy =  – 4
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = 2\\
   y =  – 3
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 4\\
   y =  – 1
   \end{array} \right.
   \end{array} \right.\\
    \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2; – 3} \right);\left( {4; – 1} \right)} \right\}
   \end{array}\)

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