Question

Buna va rog ajutor la exercitiul 3. Multumesc anticipat!!!

Answer 1

   
[tex]\displaystyle\\ \text{Se dau numerele:}~~x+2;~6;~4x+1,~unde~x\in R\\\\ a)\\ m_g=\sqrt[\b3]{(x+2)\cdot6\cdot(4x+1)}=\\\\ =\sqrt[\b3]{6\cdot(x+2)(4x+1)}=\sqrt[\b3]{6\cdot(4x^2+8x+x+2)}=\\\\ =\sqrt[\b3]{6\cdot (4x^2+9x+2)}=\boxed{\sqrt[\b3]{24x^2+54x+12}}\\\\ b)\\ \text{Rezolvam ecuatia:}\\\\ \sqrt[\b3]{24x^2+54x+12}=6~~\Big|~\text{Ridicam la puterea a 3-a.}\\\\ \left(\sqrt[\b3]{24x^2+54x+12}\right)^\Big3=6^\Big3\\\\ 24x^2+54x+12=6^3~~~\Big|~:6\\\\ 4x^2+9x+2=6^2\\ 4x^2+9x+2=36[/tex]

[tex]\displaystyle\\ 4x^2+9x+2-36=0\\ 4x^2+9x-34=0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}=\frac{-9\pm \sqrt{81+16\cdot 34} }{8}=\\\\ =\frac{-9\pm\sqrt{81+544}}{8}=\frac{-9\pm\sqrt{625}}{8}=\frac{-9\pm25}{8} \\\\ x_1 = \frac{-9+25}{8}= \frac{16}{8}=\boxed{\bf 2}\\\\ x_2 = \frac{-9-25}{8}= \frac{-34}{8}=\boxed{\bf -\frac{17}{4}}\\\\[/tex]