Tìm x,y thuộc Z biết |x-5|+|y-1|=1 Giúp mik nhanh nha! Thank

Đáp án:

\(S = \left\{ {\left( {6;\,\,1} \right),\,\,\left( {4;\,\,1} \right),\,\,\left( {5;\,\,2} \right),\,\,\left( {5;\,\,0} \right)} \right\}.\)

Giải thích các bước giải:

\(\left| {x - 5} \right| + \left| {y - 1} \right| = 1\,\,\,\left( * \right)\)

Ta có:\(\left\{ \begin{array}{l}\left| {x - 5} \right| \ge 0\,\,\,\forall x\\\left| {y - 1} \right| \ge 0\,\,\,\forall y\end{array} \right.\)

Mà \(x,\,\,y \in \mathbb{Z}\)

\(\Rightarrow \left( * \right) \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\left| {x - 5} \right| = 1\\\left| {y - 1} \right| = 0\end{array} \right.\\\left\{ \begin{array}{l}\left| {x - 5} \right| = 0\\\left| {y - 1} \right| = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\left[ \begin{array}{l}x - 5 = 1\\x - 5 =  - 1\end{array} \right.\\y - 1 = 0\end{array} \right.\\\left\{ \begin{array}{l}x - 5 = 0\\\left[ \begin{array}{l}y - 1 = 1\\y - 1 =  - 1\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\left[ \begin{array}{l}x = 6\\x = 4\end{array} \right.\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x = 5\\\left[ \begin{array}{l}y = 2\\y = 0\end{array} \right.\end{array} \right.\end{array} \right..\)

Vậy phương trình có tập nghiệm: \(S = \left\{ {\left( {6;\,\,1} \right),\,\,\left( {4;\,\,1} \right),\,\,\left( {5;\,\,2} \right),\,\,\left( {5;\,\,0} \right)} \right\}.\)