Întrebare Exercitiu multime

\(M=\left\{ x\in \mathbb{R}/\left\{ \frac{5x-2}{3} \right\}=x+1 \right\}\)

Multumesc anticipat.. as avea nevoie de cateva explicatii in rezolvarea acestei probleme!

Buna ziua,

Folosim definitia partii fractionare:

\(\left \{ x\right \}=x-[x],\: \forall x\in \mathbb{R}\), deci \(\left \{ \frac{5x-2}{3} \right \}=\frac{5x-2}{3}-\left [ \frac{5x-2}{3} \right ]\).

Atunci avem ecuatia \(\left \{ \frac{5x-2}{3} \right \}=x+1\Leftrightarrow \left [ \frac{5x-2}{3} \right ]=\frac{2x-5}{3}\).

Cum pe partea dreapta a ecuatiei avem \(\left [ \frac{5x-2}{3} \right ]\), un numar intreg,
rezulta ca si \(\frac{2x-5}{3}\) este intreg.

Notam \(\frac{2x-5}{3}=k\in \mathbb{Z}\), deci \(x=\frac{3k+5}{2}\).

Avem \(\left [ \frac{5x-2}{3} \right ]=\frac{2x-5}{3}\Leftrightarrow \left [ \frac{5\cdot \frac{3k+5}{2}-2}{3} \right ]=k\) , adica \(\left [ \frac{15k+21}{6} \right ]=k\).

Putem scrie

\(k\leq \frac{15k+21}{6}<k+1\Leftrightarrow -21\leq 9k<-15,\: k\in \mathbb{Z}\)
\(\Leftrightarrow -\frac{7}{3}\leq k<-\frac{5}{3},\: k\in \mathbb{Z}\), deci \(k=-2\).

Avem \(x=\frac{3k+5}{2}=\frac{3\cdot \left ( -2 \right )+5}{2}=-\frac{1}{2}\).

Atunci \(M=...?\)

Află cum poți scrie formule pe acest forum!

Multumesc ! O sa fiu atent sa folosesc text.