Question

Salut! Cum pot sa rezolv inecuatiile asemanatoare cu: x+1/x+2>= (mai mare sau egal)2x-1/x-1
Va rog !

Answer 1

   
$\displaystyle\\ \bold{\frac{x+1}{x+2}\geq\frac{2x-1}{x-1}}~~~~~~~x\in R-\{-2;~1\}\\\\\\ \bold{\frac{x+1}{x+2}-\frac{2x-1}{x-1}\geq 0}~~~~~~~\texttt{Aducem la acelasi numitor.}\\\\\\ \bold{\frac{(x+1)(x-1)-(2x-1)(x+2)}{(x+2)(x-1)}\geq 0}\\\\\\ \bold{\frac{(x^2-1)-(2x^2+4x -x-2)}{x^2+2x -x-2}\geq 0}\\\\\\ \bold{\frac{(x^2-1)-(2x^2+3x-2)}{x^2+x-2}\geq 0}\\\\\\ \bold{\frac{x^2-1-2x^2-3x+2}{x^2+x-2}\geq 0}\\\\\\ \bold{\frac{-x^2-3x+1}{x^2+x-2}\geq 0}$

$\displaystyle\\ \texttt{Numaratorul este pozitiv intre radacini si negativ in rest.} \\ \texttt{Numitorul este negativ intre radacini si pozitiv in rest.} \\\\ \texttt{Calculam radacinile numaratorului:} \\ \\ \bold{x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{3\pm\sqrt{9+4}}{-2}= \frac{-3\mp\sqrt{13}}{2} }\\ \\ \bold{x_1 = \frac{-3-\sqrt{13}}{2}\approx -3,30~~~~~~~x_2=\frac{-3+\sqrt{13}}{2} \approx 0,15}$

$\displaystyle\\ \bold{\frac{-x^2-3x+1}{x^2+x-2}\geq 0}\\\\\\ \texttt{Calculam radacinile numitorului:}\\\\ \bold{x_{34}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1+8}}{2}= \frac{-1\pm\sqrt{9}}{2}=\frac{-1\pm 3}{2}}\\\\ \bold{x_3 = \frac{-1- 3}{2}= \frac{-4}{2}=-2~~~~~~~x_4=\frac{-1+3}{2}=\frac{2}{2}=1}\\\\\\ \texttt{Organizarea pe axa x:} \\ \\ \bold{x_1 \ \textless \ x_3 \ \textless \ x_2 \ \textless \ x_4 }\\ \texttt{adica:}\\ \bold{-3,30 \ \textless \ -2 \ \textless \ 0,15 \ \textless \ 1}$

$\displaystyle\\ \texttt{Vezi continuarea rezolvarii in tabelul din fisierul atasat.} \\ \\ \texttt{Dupa ce vezi tabelul, concluzia finala este: } \\\\ \bold{\Longrightarrow ~~~x\in\Big[\frac{-3-\sqrt{13}}{2}, -2\Big)\bigcup \Big[\frac{-3+ \sqrt{13}}{2}, 1\Big)} \\ \\ \\ \texttt{Ambele intervale sunt deschise la dreapta deoarece:} \\ \\ ~~~~~~~~~~~~~~~\bold{x\in R-\{-2;~1\}}$