Question

Exercițiul 3 subpunctele: a, b.

Answer 1

   
[tex]3) \displaystyle \\ log_a b ~\text{este deefinit daca: } a\ \textgreater \ 0, ~a \neq 1,~b\ \textgreater \ 0. \\ \\ Rezolvare: \\ a) \\ log_5(4x^2+11x-3) \\ 4x^2+11x-3=0 \\ \\ x_{_{12}}= \frac{-11 \pm \sqrt{121+48} }{8}=\frac{-11 \pm \sqrt{169} }{8}=\frac{-11 \pm 13}{8} \\ \\ \\ x_{_1}= \frac{-11 - 13}{8} = \frac{-24}{8} = \boxed{-3} \\ \\ x_{_2}= \frac{-11 + 13}{8} = \frac{2}{8} =\boxed{ \frac{1}{4} } [/tex]

[tex]\displaystyle \\ \Longrightarrow ~~x \in \left(-\infty,~ -3~ \right) \bigcup \left( \frac{1}{4},~ \infty \right) [/tex]


[tex]b) \displaystyle \\ \frac{x-1}{x+1} \ \textgreater \ 0 ~~daca ~~x \in (-\infty,~-1) \bigcup (1,~\infty) \\ \\ \frac{x-1}{x+1} \neq 1,~~ daca~~x \in R \\ \\ 2x^2+x-3 =0 \\ \\ x_{_{12}}= \frac{-1 \pm \sqrt{1+24} }{4}=\frac{-1 \pm \sqrt{25} }{4} = \frac{-1 \pm 5 }{4}\\ \\ x_{_1}= \frac{-1 -5 }{4}= \frac{-6 }{4}=\boxed{\frac{-3 }{2}} \\ \\ x_{_2}= \frac{-1 +5 }{4}= \frac{4 }{4}=\boxed{1} ~~\Longrightarrow ~~ x \in \left(-\infty,~- \frac{3}{2} \right) \bigcup (1,~\infty ) [/tex]


[tex]\text{Cumuland toate conditiile, rezulta: } \\ \\ \Longrightarrow ~~ \boxed{x \in \left(-\infty,~- \frac{3}{2} \right) \bigcup (1,~\infty )}[/tex]