## 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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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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2 tuần 2021-07-12T10:57:04+00:00 1 Answers 1 views 0

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