Question

rezolvati in multimea z inecuatia :
1-2* \frac{2x+1}{x-3} + ( \frac{2x+1}{x-3})²>0

Answer 1

    
[tex]\displaystyle \\ 1-2\cdot \frac{2x+1}{x-3} + \left( \frac{2x+1}{x-3}\right)^2\ \textgreater \ 0 \\ \\ \left( \frac{2x+1}{x-3}\right)^2 - 2\cdot \frac{2x+1}{x-3} +1 \ \textgreater \ 0 \\ \\ Notatie: \\ \\ t = \frac{2x+1}{x-3} \\ \\ Obtinem: \\ t^2 - 2t + 1 \ \textgreater \ 0 \\ \texttt{Deoarece coeficientul lui }t^2 \texttt este \ \textgreater \ 0, } \\ \texttt{expresia: } t^2 - 2t + 1 \texttt{ este pozitiva in afara radacinilor. } [/tex]

[tex]\displaystyle {Calculam~radacinile: } \\ t^2 - 2t + 1 = (t-1)^2 \\ t_1 = 1 \\ t_2=1 \\ \Longrightarrow ~~ t^2 - 2t + 1 \ \textgreater \ 0 ~~ daca ~~t \neq 1\\ \\ Dar: ~~~t= \frac{2x+1}{x-3} \neq 1 \\ \\ \texttt{Calculam valoarea lui x pentru care fractia esde egala cu 1:} \\ 2x+1 \neq x-3 ~~\Longleftrightarrow ~~ 2x-x \neq -3-1 ~~\Longleftrightarrow ~~ \boxed{x \neq -4 }\\ \\ \Longrightarrow ~~ \boxed{x \in R - \{-4\}} [/tex]