Question

Va rog foarte mult ajutatima 

 

Answer 1

    

$a=\sqrt{2+ \sqrt{2}+ \sqrt{3+2 \sqrt{2}}}+ \sqrt{2- \sqrt{2}- \sqrt{3-2 \sqrt{2}}}= \\ =\sqrt{2+ \sqrt{2}+ \sqrt{2+2 \sqrt{2}+1}}+ \sqrt{2- \sqrt{2}- \sqrt{2-2 \sqrt{2}+1}}= \\ =\sqrt{2+ \sqrt{2}+ \sqrt{(\sqrt{2}+1)^{2}}}+ \sqrt{2- \sqrt{2}- \sqrt{(\sqrt{2}-1)^{2}}}= \\ =\sqrt{2+ \sqrt{2}+ (\sqrt{2}+1)}+ \sqrt{2- \sqrt{2}- (\sqrt{2}-1)}= \\ =\sqrt{3+2 \sqrt{2}} + \sqrt{3-2 \sqrt{2}}= \\ =\sqrt{2+2 \sqrt{2}+1} + \sqrt{2-2 \sqrt{2}+1}= \\ =\sqrt{(\sqrt{2}+1)^{2}} + \sqrt{(\sqrt{2}-1)^{2}}=$
$=(\sqrt{2}+1)+ (\sqrt{2}-1)= \\ =2\sqrt{2}$


$b=\sqrt{9-4 \sqrt{2}} + \sqrt{9+4 \sqrt{2}}= \\ =\sqrt{8-4 \sqrt{2}+1} + \sqrt{8+4 \sqrt{2}+1}= \\ = \sqrt{(2\sqrt{2}-1)^{2}} + \sqrt{(2\sqrt{2}+1)^{2}}= \\ =2\sqrt{2}-1+2\sqrt{2}+1= \\ =4 \sqrt{2}$


$m_a = \frac{2 \sqrt{2} +4 \sqrt{2} }{2}= \frac{6\sqrt{2} }{2} =2 \sqrt{2} \\ \\ m_g= \sqrt{2 \sqrt{2}*4 \sqrt{2}} =\sqrt{2 *2*4 }= \sqrt{16}=4 \\ \\ m_h= \frac{2}{ \frac{1}{2 \sqrt{2}}+ \frac{1}{4 \sqrt{2}}}= \frac{2}{\frac{2}{4 \sqrt{2}}+ \frac{1}{4 \sqrt{2}}} = \frac{2}{ \frac{2+1}{4 \sqrt{2}}}= \frac{2*4 \sqrt{2}}{3} =\frac{8 \sqrt{2}}{3}$