Lời giải:

$E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}$

$A=\left\{1; -4\right\}$

$B=\left\{-1; 2\right\}$

Do đó:

$A\cup B = \left\{-4; -1; 1;2\right\}$

$C_E(A\cup B)=\left\{-5;-3;-2; 0;3;4;5\right\}$

$A\cap B = \varnothing$

$C_E(A\cap B)=E$

\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

\(A=\left\{1;-4\right\}\)

\(B=\left\{2;-1\right\}\)

a) Với mọi x thuộc A đều thuộc E \(\Rightarrow A\subset E\)

Với mọi x thuộc B đều thuộc E \(\Rightarrow B\subset E\)

b) \(A\cap B=\varnothing\)

\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

\(A\cup B=\left\{-4;-1;1;2\right\}\)

\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)

\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)

\(A=\left\{x\in Z,x^2< 4\right\}\)

\(\Rightarrow A=\left\{-1;0;1\right\}\)

\(B=\left\{x\in Z,\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}\)\(\Rightarrow\left[{}\begin{matrix}5x-3x^2=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{3}\left(loai\right)\\x=0\\x=3\\x=-1\end{matrix}\right.\)

\(\Rightarrow B=\left\{0;-1;3\right\}\)

\(\Rightarrow A\cap B=\left\{0;-1\right\}\) \(A\cup B=\left\{0;-1;1;3\right\}\)

\(A\backslash B=\left\{1\right\}\) \(B\backslash A=\left\{3\right\}\)

\(A\cap B=\left\{{}\begin{matrix}x>m\\x\le\dfrac{2m-1}{3}\end{matrix}\right.\left(1\right)\)

 \(TH1:m< \dfrac{2m-1}{3}\)

\(\Leftrightarrow m-\dfrac{2m-1}{3}< 0\)

\(\Leftrightarrow\dfrac{m-1}{3}< 0\)

\(\Leftrightarrow m< 1\)

\(\left(1\right)\Leftrightarrow A\cap B=\left\{x\in Z|m< x\le\dfrac{2m-1}{3}\right\}\)

\(TH2:m>\dfrac{2m-1}{3}\)

\(\Leftrightarrow m-\dfrac{2m-1}{3}>0\)

\(\Leftrightarrow\dfrac{m-1}{3}>0\)

\(\Leftrightarrow m>1\)

\(\left(1\right)\Leftrightarrow A\cap B=\varnothing\)

\(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)

\(\Rightarrow A=\left(-2;4\right)\)

\(\left|x+2\right|>5\Rightarrow\left[{}\begin{matrix}x+2>5\\x+2< -5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>3\\x< -7\end{matrix}\right.\)

\(\Rightarrow B=\left(-\infty;-7\right)\cup\left(3;+\infty\right)\)

\(A\cup B=\left(-\infty;-7\right)\cup\left(-2;+\infty\right)\)

\(A\cap B=\left(3;4\right)\)

\(A\cup B=\left(-1;+\infty\right)\)

\(A\cap B=(2;5]\)

Tham khảo:

Ta có:

Bất phương trình \(1 - 2x \le 0\) có nghiệm là \(x \ge \frac{1}{2}\) hay \(A = [\frac{1}{2};+\infty)\)

Bất phương trình \(x - 2 < 0\) có nghiệm là \(x < 2\) hay \(B = ( - \infty ;2)\)

Vậy \(A \cup B = \mathbb R\)

Vậy \(A \cap B = [\frac{1}{2};2)\)

a, \(A\cup B=(-4;5]\)

\(A\cap B=[-3;4)\)

\(A\backslash B=\left[4;5\right]\)

\(B\backslash A=\left(-4;-3\right)\)

b, \(A\cup B=\left(-3;7\right)\)

\(A\cap B=[1;2)\cup(3;5]\)

\(A\backslash B=\left[2;3\right]\)

\(B\backslash A=\left(-3;1\right)\cup\left(5;7\right)\)

c, \(A\cup B=\left[\dfrac{1}{2};3\right]\)

\(A\cap B=\left[1;\dfrac{3}{2}\right]\)

\(A\backslash B=[\dfrac{1}{2};1)\)

\(B\backslash A=(\dfrac{3}{2};3]\)

d, \(A\cup B=(-5;2]\cup(3;6]\)

\(A\cap B=\left\{0\right\}\cup[4;5)\)

\(A\backslash B=(0;2]\cup\left[-5;6\right]\)

\(B\backslash A=[-5;0)\cup\left(3;4\right)\)