[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]
