\(-10\le\left|5-x\right|\le2\)

\(\Rightarrow\left|5-x\right|=1;2\)

\(\Rightarrow5-x=\pm1\) 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}

`-1/5<=x/8<=1/4`

`=>8* -1/5<=x<=1/4*8`

`=>-8/5<=x<=2`

Mà `x in ZZ`

`=>x in {-1,0,1,2}`

Mà 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)

\(2\le\left|x\right|\le5\)

\(\Rightarrow\left|x\right|\in\left\{2;3;4;5\right\}\)

\(\Rightarrow x\in\left\{\pm2;\pm3;\pm4;\pm5\right\}\)

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 \(x\in\left\{\mp2;\mp3;\mp4;\mp5\right\}\)

# HOK TỐT #

\(2\le\left|x\right|\le5\)

\(\Rightarrow\left|x\right|\in\left\{2;3;4;5\right\}\)

\(\Rightarrow x\in\left\{\pm2;\pm3;\pm4;\pm5\right\}\)

Ta có : \(-\frac{5}{6}+\frac{8}{3}+\frac{29}{-6}=-3\) và \(\frac{1}{2}+2+\frac{5}{2}=5\)

Vậy -3 < x < 5. Do x \(\in\) Z nên x \(\in\) {-2; -1; 0; 1; 2; 3; 4}

Ta có: \(A = \left\{ {x \in \mathbb{Z}| - 2 \le x \le 3} \right\} = \{  - 2; - 1;0;1;2;3\} \)

Và \(B = \{ x \in \mathbb{R}|{x^2} - x - 6 = 0\}  = \{  - 2;3\} \)

Khi đó:

Tập hợp \(A\,{\rm{\backslash }}\,B\) gồm các phần tử thuộc A mà không thuộc B. Vậy\(A\,{\rm{\backslash }}\,B = \{  - 1;0;1;2\} \).

 Tập hợp \(B\,{\rm{\backslash }}\,A\) gồm các phần tử thuộc B mà không thuộc A. Vậy \(B\,{\rm{\backslash }}\,A = \emptyset \)

a: \(x\in B\left(5\right)\)

=>\(x\in\left\{0;5;10;15;20;25;30;35;40;...\right\}\)

mà 20<=x<=36

nên \(x\in\left\{20;25;30;35\right\}\)

b: \(x\inƯ\left(20\right)\)

=>\(x\in\left\{1;2;4;5;10;20\right\}\)

mà x>8

nên \(x\in\left\{10;20\right\}\)