Question

Rezolvare completa........

Answer 1

   
[tex]\displaystyle\\ \left(\frac{9}{25}\right)^x -m\left(\frac{3}{5}\right)^x+1\ \textgreater \ 0~~~\text{Pentru oricare }~x\ \textless \ 0\\\\ \text{Inecuatia data este echivalenta cu urmatoarea inecuatie:}\\\\ \left(\frac{25}{9}\right)^x -m\left(\frac{5}{3}\right)^x+1\ \textgreater \ 0~~~\text{Pentru oricare }~x\ \textgreater \ 0\\\\ \left[\left(\frac{5}{3}\right)^2\right]^x -m\left(\frac{5}{3}\right)^x+1\ \textgreater \ 0~~~\text{Pentru oricare }~x\ \textgreater \ 0\\\\ [/tex]


[tex]\displaystyle\\ \left[\left(\frac{5}{3}\right)^x\right]^2 -m\left(\frac{5}{3}\right)^x+1\ \textgreater \ 0~~~\text{Pentru oricare }~x\ \textgreater \ 0\\\\ \text{Substitutie: }~~~\left(\frac{5}{3}\right)^x=y\\\\ y^2-my+1\ \textgreater \ 0 ~~~~~~\text{Pentru oricare }~y\ \textgreater \ 0\\\\ \text{Coeficientul lui y este pozitiv.} \\ \text{Rezulta ca functia este pozitiva in afara radacinilor.}\\\\ \Delta = b^2-4ac =(- m)^2-4\cdot 1 = m^2-4\\ \Delta \ \textless \ 0\\ m^2-4\ \textless \ 0\\ m^2 \ \textless \ 4\\ m\ \textless \ 2\\ \boxed{\bf m \in (-\infty,~2)} [/tex]