Đáp án:
$a) A=\left \{ \dfrac{n}{n^2-1}| n\in \mathbb{N},2\leq n\leq 6 \right \}\\
b)
B=\left \{ x^2-1|x\in \mathbb{N}, 1\leq x\leq 6 \right \}\\
c) C=\left \{ 5n+1|n\in \mathbb{N}, -1\leq n\leq 3 \right \}\\
d)
D=\left \{ 1-(-1)^{n+1}.3n| n\in \mathbb{N}, 0\leq n\leq 2 \right \}\\$
Giải thích các bước giải:
$a) A=\left \{ \dfrac{2}{3};\dfrac{3}{8};\dfrac{4}{15};\dfrac{5}{24};\dfrac{6}{35} \right \}\\
=\left \{ \dfrac{n}{n^2-1}| n\in \mathbb{N},2\leq n\leq 6 \right \}\\
b)
B=\{0;3;8;15;24;35\}\\
=\left \{ x^2-1|x\in \mathbb{N}, 1\leq x\leq 6 \right \}\\
c) C=\{-4;1;6;11;16\}\\
=\left \{ 5n+1|n\in \mathbb{N}, -1\leq n\leq 3 \right \}\\
d)
D=\{1;-2;7\}\\
=\left \{ 1-(-1)^{n+1}.3n| n\in \mathbb{N}, 0\leq n\leq 2 \right \}\\$
