Đáp án:
`a)` `A={-3;1}`
`b)` `A={1;3;5;7;9;11;...}`
Giải thích các bước giải:
`a) A = {x ∈ Z, (x + 3)(3x^2 – 5x + 2) = 0}`
Ta có:
`\qquad (x+3)(3x^2-5x+2)=0`
`<=>`$\left[\begin{array}{l}x+3=0\\3x^2-5x+2=0\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=-3\in Z\\x=1\in Z\\x=\dfrac{2}{3}∉Z\end{array}\right.$
`=>A={-3;1}`
$\\$
`b) A = {n ∈ NN, n = 2k + 1, k ∈ NN}`
Vì `k\in NN=>k\in {0;1;2;3;4;5;...}`
`\qquad n=2k+1`
`=>n\in {1;3;5;7;9;11;...}`
`=>A={1;3;5;7;9;11;...}`