Tìm x y thuộc n sao cho x y – 5 x + y = 17 17/07/2021 Bởi Rose Tìm x y thuộc n sao cho x y – 5 x + y = 17
Đáp án: \[\left( {x;y} \right) = \left\{ {\left( {0;17} \right);\left( {11;6} \right);\left( {1;11} \right);\left( {5;7} \right);\left( {2;9} \right);\left( {3;8} \right)} \right\}\] Giải thích các bước giải: Ta có: \(\begin{array}{l}xy – 5x + y = 17\\ \Leftrightarrow \left( {xy – 5x} \right) + \left( {y – 5} \right) = 17 – 5\\ \Leftrightarrow x\left( {y – 5} \right) + \left( {y – 5} \right) = 12\\ \Leftrightarrow \left( {x + 1} \right)\left( {y – 5} \right) = 12\\x,y \in N \Rightarrow x + 1 \ge 1\\12 = 1.12 = 2.6 = 3.4\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 1 = 1\\y – 5 = 12\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 12\\y – 5 = 1\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 2\\y – 5 = 6\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 6\\y – 5 = 2\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 3\\y – 5 = 4\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 4\\y – 5 = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 0\\y = 17\end{array} \right.\\\left\{ \begin{array}{l}x = 11\\y = 6\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = 11\end{array} \right.\\\left\{ \begin{array}{l}x = 5\\y = 7\end{array} \right.\\\left\{ \begin{array}{l}x = 2\\y = 9\end{array} \right.\\\left\{ \begin{array}{l}x = 3\\y = 8\end{array} \right.\end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left\{ {\left( {0;17} \right);\left( {11;6} \right);\left( {1;11} \right);\left( {5;7} \right);\left( {2;9} \right);\left( {3;8} \right)} \right\}\end{array}\) Bình luận
Đáp án: Trong hình
Giải thích các bước giải:
Xin lỗi vì hơi mờ.
Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( {0;17} \right);\left( {11;6} \right);\left( {1;11} \right);\left( {5;7} \right);\left( {2;9} \right);\left( {3;8} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
xy – 5x + y = 17\\
\Leftrightarrow \left( {xy – 5x} \right) + \left( {y – 5} \right) = 17 – 5\\
\Leftrightarrow x\left( {y – 5} \right) + \left( {y – 5} \right) = 12\\
\Leftrightarrow \left( {x + 1} \right)\left( {y – 5} \right) = 12\\
x,y \in N \Rightarrow x + 1 \ge 1\\
12 = 1.12 = 2.6 = 3.4\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 1\\
y – 5 = 12
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 12\\
y – 5 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 2\\
y – 5 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 6\\
y – 5 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 3\\
y – 5 = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 4\\
y – 5 = 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 0\\
y = 17
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 11\\
y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = 11
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 5\\
y = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
y = 9
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3\\
y = 8
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {0;17} \right);\left( {11;6} \right);\left( {1;11} \right);\left( {5;7} \right);\left( {2;9} \right);\left( {3;8} \right)} \right\}
\end{array}\)