Question

Media geometrica a numerelor :

Answer 1

$a=\frac{2}{\sqrt{3}+1}+4- \sqrt{3}\\ \\ a=\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}+4-\sqrt{3}\\ \\ a=\frac{2(\sqrt{3}-1)}{3-1}+4-\sqrt{3}\\ \\ a=\frac{2(\sqrt{3}-1)}{2}+4-\sqrt{3}\\ \\ a=\sqrt{3}-1+4-\sqrt{3}\\ \\ a=-1+4\\ \\ a=3\\ \\ \\ \\ b= \frac{2}{(\sqrt{2}+1)^2}+(2+\sqrt{2})^{2}\\ \\ b=\frac{2(\sqrt{2}-1)^{2}}{(\sqrt{2}+1)^{2}(\sqrt{2}-1)^{2}}+(2+\sqrt{2})^{2}\\ \\ b=\frac{2(\sqrt{2}-1)^{2}}{(\sqrt{2}-1)(\sqrt{2}+1)(\sqrt{2}-1)(\sqrt{2}+1)}+(2+\sqrt{2})^{2}\\ \\ b=\frac{2(\sqrt{2}-1)^{2}}{(2-1)(2-1)}+(2+\sqrt{2})^{2}\\ \\ b=\frac{2(\sqrt{2}-1)^{2}}{1}+(2+\sqrt{2})^{2}\\ \\ b=2(2-2\sqrt{2}+1)+(4+4\sqrt{2}+2)\\ \\ b=2(3-2\sqrt{2})+6+4\sqrt{2}\\ \\ b=6-4\sqrt{2}+6+4\sqrt{2}\\ \\ b=6+6\\ \\ b=12\\ \\ \\ \\ m_{g}=\sqrt{a\cdot b}\\ \\ m_{g}=\sqrt{3\cdot 12}\\ \\ m_{g}=\sqrt{36}\\ \\ \boxed{m_{g}=6}$