a: A={0;1;2;3}

b: B={-16;-13;-10;-7;-4;-1;2;5;8}

c: C={-9;-8;-7;...;7;8;9}

d: \(D=\varnothing\)

a: \(3< n^2< 30\)

=>\(\sqrt{3}< n< \sqrt{30}\)

mà \(n\in Z^+\)

nên \(n\in\left\{2;3;4;5\right\}\)

=>A={2;3;4;5}

b: |n|<3

=>-3<n<3

mà \(n\in Z\)

nên \(n\in\left\{-2;-1;0;1;2\right\}\)

=>B={-2;-1;0;1;2}

c: x=3k

=>\(x⋮3\)

mà -4<x<12

nên \(x\in\left\{-3;0;3;6;9\right\}\)

=>C={-3;0;3;6;9}

d: \(n\in N\)

mà n<5

nên \(n\in\left\{0;1;2;3;4\right\}\)

=>\(n^2+3\in\left\{3;4;7;12;19\right\}\)

=>D={3;4;7;12;19}

\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)

\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)

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ví dụ là 3k + 1 = 3 . 4 + 1 = 13 

13 khi chia cho 3 thì còn dư 1  3k + 2 cũng vậy , 2 là số dư của phép tính đó  

Oki, thank you nha!
CHÚC BẠN THI GIỮA KÌ TỐT

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) \(C=\left\{-3;3;6;9\right\}\)

Ta thấy 3k+1 là số chẵn, 6m+1 là số lẻ với \(k,m\ne0\). Với k=m=0: 3k+1=6m+1=1.

Vậy \(A\cap B=\left\{1\right\}\);A\B={3k+1|\(k\in\text{ℕ*}\)}

#Walker