Question

Ajutir am nevoie urgent !!!!​

Answer 1

5.

$\it x^2-2x-3 =x^2-2x+1-4=(x-1)^2-2^2=\\ \\ =(x-1-2)(x-1+2)=(x-3)(x+1)=(x+1)(x-3)\\ \\ (x+1)(x-3)=0 \Rightarrow x_1=-1, \ x_2=3 \Rightarrow A=\{-1,\ 3\}\\ \\ \\ 3x-6\geq0 \Rightarrow 3x\geq6|_{:3} \Rightarrow x\geq2 \Rightarrow B=[2,\ \infty)$

$\it A\cap B=\{3\}\\ \\ A\cup B=\{-1\}\cup[2,\ \infty)\\ \\ B\setminus A=[2,\ \ 3)\cup(3,\ \infty)$

4.

$\it x^2+2x-3=0 \Rightarrow x^2+3x-x-3=0 \Rightarrow x(x+3)-(x+3)=0 \Rightarrow \\ \\ \Rightarrow (x+3)(x-1)=0 \Rightarrow x_1=-3,\ \ x_2=1 \Rightarrow S_1=\{-3,\ \ 1\}\\ \\ (x-3)(x^3-1)=0 \Rightarrow \begin{cases} \it x^3-1=0 \Rightarrow x^3=1 \Rightarrow x_1=1\\ \\ \it x-3=0 \Rightarrow x_2=3\end{cases} \Rightarrow S_2=\{1,\ \ 3\}$

$\it S_1\cap S_2=\{1\}\\ \\ S_1\cup S_2=\{-3,\ \ 1,\ \ 3\}$