Question

determinati valorile naturale ale lui x pentru care x+1 este divizor al numarului (x la a doua)+5

Answer 1

   
[tex]\displaystyle \\ (x+1) ~\texttt{este divizor al lui }~(x^2+5)~~\texttt{este echivalent cu:}\\\\ \frac{x^2+5}{x+1} \in N \\\\ \texttt{Rezolvare: } \\\\ \frac{x^2+5}{x+1}=\frac{x^2-1+6}{x+1}=\frac{x^2-1}{x+1} + \frac{6}{x+1} = x-1 + \frac{6}{x+1} \\\\ D_6 =\{ 1;~2;~3;~6\} \\\\ x+1 = 1~~ \Longrightarrow~~ x=0 \\\\ x+1 = 2~~ \Longrightarrow~~ x=1 \\\\ x+1 = 3~~ \Longrightarrow~~ x=2 \\\\ x+1 = 6~~ \Longrightarrow~~ x=5 \\\\ \texttt{Raspuns: }~~~\boxed{x\in \{0;~1;~2;~6\}}[/tex]