Question

$a)Aflati-l ~pe~x,stiind~ca: \\ parte~intreaga~(x)+parte~intreaga~(5x+ \frac{2}{3} )=x \\b)Aflati-l ~pe~x,stiind~ca : \\ parte~fractionara~x + parte ~fractionara~ \frac{5x+3}{x+3} =x \\ x~apartine~lui~R~in~ambele~cerinte!$

Answer 1

$a)~[x]+\Big[5x+ \frac{2}{3} \Big]=x. \\ \\ Se~constata~ca~x \in Z~(suma~de~parti~intregi),~de~unde~[x]=x, \\ \\ de~unde~rezulta~ \Big[5x+ \frac{2}{3} \Big]=0.~(De~asemenea~se~putea~trece \\ \\ \ [x]~in~membrul~drept,~si~rezulta ~\Big[5x+ \frac{2}{3} \Big]= \{x\}). \\ \\ \Big[5x+ \frac{2}{3} \Big]=0 \Rightarrow 0 \leq 5x+ \frac{2}{3}\ \textless \ 1 \Rightarrow - \frac{2}{15} \leq x\ \textless \ \frac{1}{15},~dar~x \in Z \Rightarrow \\ \\ \Rightarrow \boxed{x=0}~.$

$b)~\{x \}+ \Big \{ \frac{5x+3}{x+3} \Big \}=x \Leftrightarrow \Big \{ \frac{5x+3}{x+3} \Big \} =x- \{ x \}. \Leftrightarrow \\ \\ \Leftrightarrow \Big \{ \frac{5x+3}{x+3} \Big \}= [x] \Rightarrow \Big \{ \frac{5x+3}{x+3} \Big \}=[x]=0. \\ \\ Deci~x \in [0;1)~si~ \frac{5x+3}{x+3} =k \in Z. \\ \\ 5x+3=kx+3k \Leftrightarrow (5-k)x=3k-3.~Se~constata~ca~k \neq 5, \\ \\ si~avem~x= \frac{3k-3}{5-k} \in [0;1). \\ \\ Din~x \in[0;1)~rezulta~si~k\ \textgreater \ 0,~si~cum~k \in Z,~rezulta~k \geq 1.$

$3k-3 \geq 0 \Rightarrow 5-k\ \textgreater \ 0 . \\ \\ Deci~k \in \{1;2;3;4 \}. \\ \\ k=1 \Rightarrow \boxed{ x=0}. \\ \\ k=2 \Rightarrow x=1,~nu~convine. \\ \\ k=3 \Rightarrow x=3,~nu~convine. \\ \\ k=4 \Rightarrow x=9,~nu~convine.$