Question

Ex 1,multumesc! Dau coroana

Answer 1

b) 

[tex] 3[a] + 2 = 4a \Rightarrow [a] = \dfrac{4a-2}{3} \in\mathbb{Z} \\\;\\\ \\\;\\ Fie\ \ \ k \in \mathbb{Z} \ astfel \ \^{i}nc\hat{a}t\ \ \dfrac{4a-2}{3} = k\Rightarrow a=\dfrac{3k+2}{4}\ \ \ (1)[/tex]

Ecuația se scrie:


[tex]\it \left[\dfrac{3k+2}{4}\right]=k \Rightarrow k \leq \dfrac{3k+2}{4} \ \textless \ k+1|_{\cdot4} \Rightarrow 4k \leq3k+2\ \textless \ 4k+4\ \ \ (2) \\\;\\ \\\;\\ (2)\Rightarrow \begin{cases} \it4k\leq3k+2\Rightarrow k\leq2 \\\;\\ \it4k+4\ \textgreater \ 3k+2\Rightarrow k\ \textgreater \ -2\end{cases} \Longrightarrow k \in\{-1, 0, 1, 2\} \ \ \ (3)[/tex]

$\it (1), (3) \Rightarrow a \in \left\{-\dfrac{1}{4},\ \dfrac{1}{2},\ \dfrac{5}{4},\ 2} \right\}$