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\(-10\le\left|5-x\right|\le2\)
\(\Rightarrow\left|5-x\right|=1;2\)
\(\Rightarrow5-x=\pm1\) hoặc \(5-x=\pm2\)
+) \(5-x=1\Rightarrow x=4\)
+) \(5-x=-1\Rightarrow x=6\)
+) \(5-x=2\Rightarrow x=3\)
+) \(5-x=-2\Rightarrow x=7\)
Vậy \(x\in\left\{4;6;3;7\right\}\)
\(-10\le\left|5-x\right|\le2\\ \Rightarrow\left|5-x\right|=1;2\\ \Rightarrow5-x=\pm1ho\text{ặ}c5-x=\pm2\\ \)
+) \(5-x=1\\ \Rightarrow x=4\\ \)
+) \(5-x=-1\Rightarrow x=6\)
+) \(5-x=2\Rightarrow x=3\)
+) \(5-x=-2\Rightarrow x=7\\ \)
Vậy \(x\in\left\{4;6;3;7\right\}\)
\(0\le\left|x\right|\le3\) \(0\le\left|y\right|\le5\) \(x-y=2\)
Vì \(x-y=2\Rightarrow x=y+2\)\(\Rightarrow0\le\left|y+2\right|\le3\Rightarrow0\le\left|y\right|\le1\)
\(\Rightarrow\left|y\right|=\orbr{\begin{cases}1\\0\end{cases}}\)\(\Rightarrow y=\orbr{\begin{cases}\orbr{\begin{cases}1\\-1\end{cases}}\\0\end{cases}\Rightarrow x=\orbr{\begin{cases}\orbr{\begin{cases}4\\2\end{cases}}\\3\end{cases}}}\)\(\Rightarrow y=\left(-1;0;1\right)\Rightarrow x=\left(1;2;3\right)\)
\(\left(x;y\right)=\left(-1;1\right),\left(0;2\right),\left(1;3\right)\)
\(\hept{\begin{cases}!x!\le3\\!y!\le5\\x-y=2\end{cases}}\Rightarrow\hept{\begin{cases}-3\le x\le3\\-5\le y\le5\\y=x+2\end{cases}}\)
với x={-3,-2,-1,0,1,2,3}
=> y={-1,0,1,2,4,5}
\(2\le\left|x\right|\le5\)
\(\Rightarrow\left|x\right|\in\left\{2;3;4;5\right\}\)
\(\Rightarrow x\in\left\{\pm2;\pm3;\pm4;\pm5\right\}\)
Vì \(2\le|x|\le5\)mà x \(\in\)Z
\(\Rightarrow|x|\in\left\{2;3;4;5\right\}\)
\(\Rightarrow x\in\left\{\mp2;\mp3;\mp4;\mp5\right\}\)
Vậy \(x\in\left\{\mp2;\mp3;\mp4;\mp5\right\}\)
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