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

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    0
    2021-07-12T10:58:41+00:00

    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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