bună ziua! mă poate ajuta va rog cineva la acest ex?​

[tex]\it a=\dfrac{\sqrt{27}-3}{\sqrt3}=\dfrac{\sqrt3(\sqrt9-\sqrt3)}{\sqrt3}=3-\sqrt3\\ \\ \\ b=18+3\sqrt{12}=18+3\sqrt{4\cdot3}=18+6\sqrt3=6(3+\sqrt3)\\ \\ \\ m_g=\sqrt{a\cdot b}=\sqrt{(3-\sqrt3)\cdot6\cdot(3+\sqrt3)}=\sqrt{6(9-3)}=\sqrt{36}=6[/tex]

Răspuns:

b) Alesia

Explicație pas cu pas:

[tex]a = \frac{^{ \sqrt{3})} \sqrt{27} - 3 }{ \sqrt{3} } = \frac{ \sqrt{81} - 3 \sqrt{3} }{3} = \frac{9 - 3 \sqrt{3} }{3} = \frac{ \not3(3 - \sqrt{3})}{ \not3} = 3 - \sqrt{3} [/tex]

[tex]b = 18 + 3 \sqrt{12} = 18 + 6 \sqrt{3} [/tex]

[tex]m_{g} = \sqrt{ab} = \sqrt{(3 - \sqrt{3} )(18 + 6 \sqrt{3} )} = \\ = \sqrt{54 + 18 \sqrt{3} - 18 \sqrt{3} - 18} = \sqrt{36} = \bf 6[/tex]