a: A={x\(\in R\)|x^2+x-6=0 hoặc 3x^2-10x+8=0}

=>x^2+x-6=0 hoặc 3x^2-10x+8=0

=>(x+3)(x-2)=0 hoặc (x-2)(3x-4)=0

=>\(x\in\left\{-3;2;\dfrac{4}{3}\right\}\)

=>A={-3;2;4/3}

B={x\(\in\)R|x^2-2x-2=0 hoặc 2x^2-7x+6=0}

=>x^2-2x-2=0 hoặc 2x^2-7x+6=0

=>\(x\in\left\{1+\sqrt{3};1-\sqrt{3};2;\dfrac{3}{2}\right\}\)

=>\(B=\left\{1+\sqrt{3};1-\sqrt{3};2;\dfrac{3}{2}\right\}\)

A={-3;2;4/3}

b: \(B\subset X;X\subset A\)

=>\(B\subset A\)(vô lý)

Vậy: KHông có tập hợp X thỏa mãn đề bài

a) \(2x^3-3x^2-5x=0\)

\(x\left(x+1\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-1\left(TM\right)\\x=\dfrac{5}{2}\left(L\right)\end{matrix}\right.\)

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

b) \(x< \left|3\right|\)\(\Leftrightarrow-3< x< 3\)

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

c) \(C=\left\{-3;3;6;9\right\}\)

a) \(A=\left\{x\in Z|2x^3-3x^2-5x=0\right\}\)

\(2x^3-3x^2-5x=0\)

\(\Leftrightarrow x\left(2x^2-3x-5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(2x-5\right)=0\)

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

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

b) \(B=\left\{-2;-1;0;1;2\right\}\)

c) \(C=\left\{-3;3;6;9\right\}\)

`#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)(2x^2-5x+3)(x^2-4x+3)=0`

`<=>[(2x^2-5x+3=0),(x^2-4x+3=0):}<=>[(x=3/2),(x=1),(x=3):}`

  `=>A={3/2;1;3}`

`b)(x^2-10x+21)(x^3-x)=0`

`<=>[(x^2-10x+21=0),(x^3-x=0):}<=>[(x=7),(x=3),(x=0),(x=+-1):}`

   `=>B={0;+-1;3;7}`

`c)(6x^2-7x+1)(x^2-5x+6)=0`

`<=>[(6x^2-7x+1=0),(x^2-5x+6=0):}<=>[(x=1),(x=1/6),(x=2),(x=3):}`

    `=>C={1;1/6;2;3}`

`d)2x^2-5x+3=0<=>[(x=1),(x=3/2):}`   Mà `x in Z`

    `=>D={1}`

`e){(x+3 < 4+2x),(5x-3 < 4x-1):}<=>{(x > -1),(x < 2):}<=>-1 < x < 2`

    Mà `x in N`

   `=>E={0;1}`

`f)|x+2| <= 1<=>-1 <= x+2 <= 1<=>-3 <= x <= -1`

      Mà `x in Z`

  `=>F={-3;-2;-1}`

`g)x < 5`  Mà `x in N`

   `=>G={0;1;2;3;4}`

`h)x^2+x+3=0` (Vô nghiệm)

   `=>H=\emptyset`.

x[x+1][x+2]=x3+1+2=x3+3=0

suy ra x3=0-3=-3

suy ra x=-3 chia cho3

suy ra x= -1

     vậy x =-1

nhớ dùng kí hiệu nha cu

(Bấm máy tính tìm nghiệm)

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

\(B=\left\{0;1;2;3\right\}\)

a: \(A=\left\{0;1;2;3;4;5\right\}\)

b: \(B=\left\{2;3;4;5\right\}\)

c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)

giúp MK nhó các bạn 

Bài 2: 

a: 

A={0;1;2;3;4;5}

B={3;4;5;6}

b: 6

c: 0;1;2

d: 3;4;5