Question

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Answer 1

$\it |2x-3|\leq5 \Rightarrow -5\leq2x-3\leq5\Big|_{+3} \Rightarrow -2\leq2x\leq8\Big|_{:2} \Rightarrow -1\leq x\leq4 \Rightarrow \\ \\ \Rightarrow A=\{-1,\ 0,\ 1,\ 2,\ 3,\ 4\}\\ \\ \\ \dfrac{x-2}{x+2}\in\mathbb{Z} \Rightarrow x+2\Big|x-2\ \ \ \ \ (1)\\ \\ \\ Dar,\ \ x+2\big| x+2\ \ \ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow x+2\Big|(x+2)-(x+2) \Rightarrow x+2\Big|x+2-x+2 \Rightarrow x+2\Big|4 \Rightarrow$

$\it \Rightarrow x+2\in\{\pm1,\ \pm2,\ \pm4\} \Rightarrow x+2\in\{-4,\ -2,\ -1,\ 1,\ 2,\ 4\}\Big|_{-2} \Rightarrow \\ \\ \Rightarrow B=\{-6,\ -4,\ -3,\ -1,\ 0,\ 2\}\\ \\ A\cap B=\{-1,\ \ 0,\ \ 2\}\\ \\ A\setminus B=\{-6,\ -4,\ -3\}$

Answer 2

Răspuns:

Explicație pas cu pas: