Verified answer

Vom explicita modulul:

[tex]g) \ G = \bigg\{x\in\Bbb{Q} \ \Big| \ \bigg|\dfrac{3x+5}{2} = 4 \bigg|=4\bigg\}[/tex]

[tex]\dfrac{3x+5}{2} = - 4 \Rightarrow 3x+5=-8 \Rightarrow 3x=-8-5 \Rightarrow 3x=-13 \ \Big|:3\\[/tex]

[tex]\Rightarrow x=-\dfrac{13}{3} \in\Bbb{Q}[/tex]

[tex]\dfrac{3x+5}{2} = 4 \Rightarrow 3x+5=8 \Rightarrow 3x=8-5 \Rightarrow 3x=3 \ \Big|:3\\[/tex]

[tex]\Rightarrow x=1 \in\Bbb{Q}[/tex]

[tex]\Rightarrow \boldsymbol{G = \bigg\{-\dfrac{13}{3}; \ 1\bigg\}}[/tex]

[tex]h) \ H = \Big\{x\in\Bbb{Z} \Big| \ \big|2x-3\big| \leq 9 \Big\}[/tex]

[tex]- 9 \leq 2x-3 \leq 9 \ \ \Big|+3 \Rightarrow -6 \leq 2x \leq 12 \ \ \Big|:2\\[/tex]

[tex]\Rightarrow -3 \leq x \leq 6[/tex]

[tex]x \in \Bbb{Z} \Rightarrow \boldsymbol{H = \{-3,-2,-1,0,1,2,3,4,5,6\}}[/tex]

______

Reținem:

✍ Modulul reprezintă valoarea absolută a unui număr real.

[tex]\boldsymbol{| \ x \ | = \begin{cases} \ \ x, \ \ \ dac\u a \ \ x \geq 0\\-x, \ \ \ dac\u a \ \ x < 0 \end{cases}}[/tex]