Question

help................................dau coroana.......................... pls.​

Answer 1

Răspuns:

Explicație pas cu pas:

$\bf x^{2} \leq 25\Rightarrow x^{2} \leq 5^{2} \Rightarrow x \leq 5$

$\bf dar ~x \in\mathbb{N^{\star}} \Rightarrow x \in \{1,2,3,4,5\}$

$\green{\bf A =\{1,2,3,4,5\}}$

$\bf x^{3} \leq 27\Rightarrow x^{3} \leq 3^{3} \Rightarrow x \leq 3$

$\bf dar ~x \in\mathbb{N^{\star}} \Rightarrow x \in \{1,2,3\}$

$\blue{\bf B= \{1,2,3\}}$

$\bf \red{A \cup B}=\{1,2,3,4,5\}\cup\{1,2,3\}=\red{\{1,2,3,4,5\}}$

$\bf \red{A \cap B}=\{1,2,3,4,5\} \cap\{1,2,3\}=\red{\{1,2,3\}}$

$\bf \red{A \setminus B}=\{1,2,3,4,5\} \setminus \{1,2,3\}=\red{\{4,5\}}$

$\bf \red{B \setminus A}=\{1,2,3\} \setminus \{1,2,3,4,5\}=\red{\{\varnothing\}}$

$==pav38==$

Answer 2

$\it x\in\mathbb{N}^*,\ \ x^2\leq25 \Rightarrow x\in\{1,\ \ 2,\ \ 3,\ \ 4,\ \ 5\}=A\\ \\ x\in\mathbb{N}^*,\ \ x^3\leq27 \Rightarrow x\in\{1,\ \ 2,\ \ 3 \}=B\\ \\ \\ A\cup B=\{1,\ \ 2,\ \ 3,\ \ 4,\ \ 5\};\ \ A\cap B=\{1,\ \ 2,\ \ 3\}\\ \\ A\setminus B=\{4,\ \ 5\};\ \ \ B\setminus A= \varnothing$