Question

PLSSS VREAU EX 4 PUNCTELE B,C SI D!!!!!!​

Answer 1

Explicație pas cu pas:

4.b)

$\Big|\dfrac{n + 1}{2n + 3} - \dfrac{1}{2}\Big| > \dfrac{1}{12}$

I)

$\dfrac{n + 1}{2n + 3} - \dfrac{1}{2} < - \dfrac{1}{12} \\ \dfrac{n + 1}{2n + 3} - \dfrac{1}{2} + \dfrac{1}{12} < 0$

$\dfrac{n + 1}{2n + 3} - \dfrac{5}{12} < 0 \\ \dfrac{12(n + 1) - 5(2n + 3)}{12(2n + 3)} < 0 \\ \dfrac{12n + 12 - 10n - 15}{12(2n + 3)} < 0 \\ \dfrac{2n - 3}{12(2n + 3)} < 0$

$2n - 3 = 0 \implies n = \dfrac{3}{2} \\2n + 3 = 0 \implies n = - \dfrac{3}{2}$

$n \in \Big(- \dfrac{3}{2} ; \dfrac{3}{2} \Big) \cap \mathbb{Z} \implies n \in \Big\{ - 1; 0; 1 \Big\}$

II)

$\dfrac{n + 1}{2n + 3} - \dfrac{1}{2} > \dfrac{1}{12} \\ \dfrac{n + 1}{2n + 3} - \dfrac{7}{12} > 0 \\ \dfrac{12(n + 1) - 7(2n + 3)}{12(2n + 3)} > 0 \\ \dfrac{12n + 12 - 14n - 21}{12(2n + 3)} > 0 \\ \dfrac{ - 2n - 9}{12(2n + 3)} > 0 \iff \dfrac{2n + 9}{12(2n + 3)} < 0$

$2n + 9 = 0 \implies n = - \dfrac{9}{2} \\ 2n + 3 = 0 \implies n = - \dfrac{3}{2}$

$n \in \Big(- \dfrac{9}{2} ; - \dfrac{3}{2} \Big) \cap \mathbb{Z} \implies n \in \Big\{ - 4; - 3; - 2 \Big\}$

=>

$n \in \Big\{- 1; 0; 1 \Big\} \ \cup \Big\{ - 4; - 3; - 2 \Big\} \\ \implies n \in \Big\{- 4; - 3; - 2;- 1; 0; 1 \Big\}$