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Media geometrica a numerelor : (2radical3 +1)la puterea a doua , 13-4radical3

Răspuns :

Media geometrica = [tex]\bf \sqrt{a\cdot b}[/tex] , a,b ≥ 0

Numerele: [tex]\bf (2\sqrt{3}+1)^{2}[/tex]   si    [tex]\bf 13-4\sqrt{3}[/tex]

[tex]\bf Mg = \sqrt{(2\sqrt{3}+1)^{2}\cdot(13-4\sqrt{3} )}[/tex]

[tex]\bf Mg = \sqrt{(2\sqrt{3})^{2}+2\cdot2\sqrt{3} +1)\cdot(13-4\sqrt{3} )}[/tex]

[tex]\bf Mg = \sqrt{(4\cdot3+4\sqrt{3} +1)\cdot(13-4\sqrt{3} )}[/tex]

[tex]\bf Mg = \sqrt{(12+4\sqrt{3} +1)\cdot(13-4\sqrt{3} )}[/tex]

[tex]\bf Mg = \sqrt{(13+4\sqrt{3})\cdot(13-4\sqrt{3} )}[/tex]

[tex]\bf Mg = \sqrt{13^{2}-(4\sqrt{3})^{2}}[/tex]

[tex]\bf Mg = \sqrt{169-4^{2}\cdot(\sqrt{3})^{2}}[/tex]

[tex]\bf Mg = \sqrt{169-16\cdot3}}[/tex]

[tex]\bf Mg = \sqrt{169-48}[/tex]

[tex]\bf Mg = \sqrt{121}[/tex]

[tex]\bf Mg = \sqrt{11^{2}}[/tex]

[tex]\boxed{\bf Mg = 11}[/tex]

Raspuns: 11 este media geometrica a numerelor