Question

1.Rezolvati in Z ecuatia:
a)|x|=5
b)|x-5|=0
c)|x+4|=2
d)|x|=0
e)|x+6|=2
f)|x-2|=-1
g)|3x+1|=10
h) |3x+4|=19
i)|4x+10|=-3
2.Determinati A∪B stiind ca A={x∈Z/4x-3>5} si B={x∈Z/|x|≤6}
va rog,dau coroana

Answer 1

$1) \\ \\ a) \quad |x| = 5 \Rightarrow \left\| \begin{array}{c} x=-5 \\ sau \\ x=5 \end{array} \right| \Rightarrow \boxed{x \in \Big\{-5,5\Big\}} \\ \\ b)\quad |x-5| = 0 \Rightarrow \left\| \begin{array}{c} x-5= -0 \\ sau \\x-5= +0 \end{array} \right |\Rightarrow x-5 = 0 \Rightarrow \boxed{x = 5}$

$c) \quad |x+4| = 2 \Rightarrow \left\| \begin{array}{c} x+4 = -2 \\ sau \\ x+4 = 2 \end{array} \right \Rightarrow \left\| \begin{array}{c} x = -2-4 \\ sau \\ x = 2-4 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\| \begin{array}{c} x= -6 \\ sau \\ x = -2 \end{array} \right | \Rightarrow \boxed{x\in\Big\{-6,-2\Big\}}$

$f) \quad \underset{\big{\geq0}}{|x-2|} = -1 (F) \Rightarrow \boxed{x\in \emptyset} \\ \\ g) \quad |3x+1| = 10 \Rightarrow \left\| \begin{array}{c}3x+1= -10 \\ sau \\ 3x+1 = 10 \end{array} \right \Rightarrow \left\| \begin{array}{c} 3x = -11 \\ 3x = 9 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\| \begin{array}{c} x = \dfrac{-11}{3} \notin\mathbb_{Z} \\ x = 3 \end{array} \right |\Rightarrow \boxed{x=3}$

$i) \quad \underset{\big{\geq 0}}{|4x+10|} = -3 (F) \Rightarrow \boxed{x \in \emptyset }$


$2)\\ A = \Big\{x\in \mathbb_{Z} \big| 4x-3\ \textgreater \ 5 \Big\} \\ B = \Big\{x\in \mathbb_{Z} \big| |x|\leq 6 \Big\} \\ \\ A: \\ 4x-3\ \textgreater \ 5 \Rightarrow 4x \ \textgreater \ 8 \Rightarrow x \ \textgreater \ 2 \Rightarrow x \in (2,\infty)\cap \mathbb{Z} \Rightarrow x\in \Big\{3,4,5,...,\infty\Big\}\\ \\ B:\\ |x|\leq 6 \Rightarrow -6\leq x\leq 6 \Rightarrow x\in [-6,6]\cap \mathbb_{Z} \Rightarrow \\ \\ \Rightarrow x \in\Big\{-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6\Big\} \\ \\ \\ \Rightarrow \boxed{A\cup B = \Big\{-6,-5,-4,-3,-2,-1,0,1,...,\infty\Big\}}$