[tex]11.~Domeniul~de~de.fi.ni.ti.e:~2x+1 \geq 0 \Leftrightarrow x \geq - \frac{1}{2}. \\ \\ Vom~analiza~doua~cazuri: \\ \\ i)~x \in [- \frac{1}{2};3).~In~acest~caz~3x-9 \ \textless \ 0,~dar~ \sqrt{2x+1} \geq 0,~ \\ \\ rezulta~ \sqrt{2x+1} \ \textgreater \ 3x-9,~deci~[- \frac{1}{2};3)~este~o~solutie . \\ \\ ii)~x \in [3;+ \infty).~In~acest~caz,~ambii~membri~sunt~ \geq 0,~deci \\ \\ putem~ridica~inecuatia~la~patrat.~Obtinem: [/tex]
[tex]2x+1\ \textgreater \ 9x^2-54x+81 \Leftrightarrow 0\ \textgreater \ 9x^2-56x+80 \Leftrightarrow \\ \\ \Leftrightarrow (9x-20)(x-4)\ \textless \ 0.~Deoarece~x \geq 3 \Rightarrow 9x-20\ \textgreater \ 0 \Rightarrow \\ \\ \Rightarrow x-4\ \textless \ 0 \Rightarrow x\ \textless \ 4,~si~cum~x \geq 3~obtinem~a~doua~solutie \\ \\ intervalul ~[3;4). \\ \\ Solutie:~[ -\frac{1}{2};3) \cup [3;4) = \boxed{\Big[- \frac{1}{2};4 \Big) }. [/tex]
[tex]27.~Domeniul~de~de.fi.ni.ti.e:~x^2+4x-5 \geq 0 \Leftrightarrow \\ \\ \Leftrightarrow (x-1)(x+5) \geq 0 \Leftrightarrow x \leq -5~sau~x \geq 1. \\ \\ Analizam~doua~cazuri: \\ \\ i)~ x \leq -5 \Rightarrow x-3\ \textless \ 0,~si~cum~ \sqrt{x^2+4x-5} \geq 0,~rezulta~ca \\ \\ intervalul~[- \infty;-5]~este~solutie. \\ \\ ii)~x \geq 1.~Partionam~acest~caz~in~doua~subcazuri: \\ \\ a) x \in [1;3] \Rightarrow x-3 \leq 0,~si~cum~ \sqrt{x^2+4x-5} \geq 0,~rezulta~ca \\ \\ intervalul~[1;3]~este~solutie. [/tex]
[tex]b)~x \in (3;+ \infty).~In~acest~caz~ambii~membri~sunt~pozitivi,~deci \\ \\ putem~ridica~inecuatia~la~patrat.~Obtinem: \\ \\ x^2-6x+9\ \textless \ x^2+4x-5 \Leftrightarrow 14\ \textless \ 10x \Rightarrow x\ \textgreater \ \frac{7}{5},~ceea~ce~este \\ \\ adevarat,~deoarece~am~presupus~x\ \textgreater \ 3. \\ \\ Deci~si~intervalul~(3;+ \infty)~este~solutie.~Intersectand~cele \\ \\ trei~solutii,~obtinem:~Solutie: \boxed{(-\infty;-5] \cup [1;+ \infty)} .[/tex]
