Question

Media geometrica a numerelor:
a este sus si b este jos

Answer 1

√7 +4√3 = √ (2+ √3)² =2 +√3

(Mg)² = [1/(2 - √3) + 2 + √3 ] · [ 2-√3  + 1/(2+√3) ] = (2-√3 )/ (2-√3) + 4-3 +
 1/(4-3) + (2+√3)/(2+√3) = 1+1 + 1 +1 =4
Mg  = √4 = 2

Answer 2

   
[tex]\displaystyle\\ a = \frac{1}{2- \sqrt{3}}+ \sqrt{7+4 \sqrt{3} }=\\\\ = \frac{1(2+ \sqrt{3})}{(2- \sqrt{3})(2+ \sqrt{3})}+ \sqrt{4+3+2\times 2\times \sqrt{3} }=\\\\ = \frac{2+ \sqrt{3}}{2^2- (\sqrt{3})^2}+ \sqrt{4+2\times 2\times \sqrt{3}+3 }=\\\\ = \frac{2+ \sqrt{3}}{4- 3}+ \sqrt{2^2+2\times 2\times \sqrt{3}+ (\sqrt{3})^2 }=\\\\ = \frac{2+ \sqrt{3}}{1}+ \sqrt{(2+\sqrt{3})^2 }=(2+ \sqrt{3}) + (2+ \sqrt{3}) = \boxed{2(2+ \sqrt{3})} [/tex]


[tex]\displaystyle\\ b = 2- \sqrt{3}+ \frac{1}{2+ \sqrt{3}} = \\\\ = 2- \sqrt{3}+ \frac{1(2- \sqrt{3})}{(2+ \sqrt{3})(2- \sqrt{3})} =\\\\ = 2- \sqrt{3}+ \frac{2- \sqrt{3}}{2^2- (\sqrt{3})^2} =\\\\ = 2- \sqrt{3}+ \frac{2- \sqrt{3}}{4- 3} =\\\\ =2- \sqrt{3}+ \frac{2- \sqrt{3}}{1} = (2- \sqrt{3})+(2- \sqrt{3})=\boxed{2(2- \sqrt{3})} [/tex]


[tex]\displaystyle\\ m_g = \sqrt{a\times b} = \sqrt{2(2+ \sqrt{3}) \times 2(2- \sqrt{3})} =\\\\ = \sqrt{2^2 (2+ \sqrt{3})(2- \sqrt{3})} =\sqrt{2^2 \times(2^2- (\sqrt{3})^2)} =\\\\ =\sqrt{4 \times(4- 3)} =\sqrt{4 \times 1} =\sqrt{4} = \boxed{\boxed{2}}[/tex]