B = { 5 ; 6 }

    = { x | \(5\le x\le6\)}

a: A={1;2;3;4}

b: B={6;7;8;9;10}

a)A=(1,2,3,4)

b)B=(6,7,8,9,10)

M = {1; 2; 3; 4; 6; 8; 12; 24}

N = {0; 7; 14; 21; 28; 35; 42; 49}

`#3107.101107`

a,

\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)

`<=> (2x - x^2)(3x - 2) = 0`

`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x\left(2-x\right)=0\\3x=2\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy, `A = {0; 2; 2/3}`

b,

\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)

`<=> 2x^3 - 3x^2 - 5x = 0`

`<=> x(2x^2 - 3x - 5) = 0`

`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

Vậy, `B = {-5/2; 0; 1}.`

c,

\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)

`<=> 2x^2 - 75x - 77 = 0`

`<=> 2x^2 - 2x + 77x - 77 = 0`

`<=> (2x^2 - 2x) + (77x - 77) = 0`

`<=> 2x(x - 1) + 77(x - 1) = 0`

`<=> (2x + 77)(x - 1) = 0`

`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)

Vậy, `C = {-77/2; 1}`

d,

\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)

`<=> (x^2 - x - 2)(x^2 - 9) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)

Vậy, `D = {-1; -3; 2; 3}.`

A={18;19;20}

a) B={23;24;25;26;27;28;29;30;31}

b) C={4;5;6;7;8;9;10}

c) D={7;8;9}

d) E={8}

\(a.E=\left\{8;9;10;11\right\}\)

\(b.A=\left\{0;30;60;90\right\}\)

a [8,9.10,11]

e =[8,9,10,11]

b= [0,30,60,90]