Question

Ma poate ajuta cineva la exercițiul 2?
Multumesc anticipat!

Answer 1

Răspuns:

$a=\sqrt{6-\sqrt{11}}~~si~~b=\sqrt{6+\sqrt{11}}\\a)m_g=\sqrt{a\cdot b}=\sqrt{(\sqrt{6-\sqrt{11}})(\sqrt{6+\sqrt{11}})}=\sqrt{\sqrt{(6-\sqrt{11})(6+\sqrt{11})}}=\\=\sqrt{\sqrt{36-11}}=\sqrt{\sqrt{25}}=\sqrt{5}$

$b)(a+b)^2+1=a^2+2ab+b^2+1=a^2+2(\sqrt{a\cdot b})^2+b^2+1=\\(\sqrt{6-\sqrt{11}})^2+2(\sqrt5)^2+(\sqrt{6+\sqrt{11}})^2+1=\\=6-\sqrt{11}+2\cdot 5+6+\sqrt{11}+1=13+10=23\\\texttt{Evident, 23 este numar prim.}$