Предмет: Алгебра, автор: vuqarnadirovkar

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Автор ответа: NNNLLL54

$A=\sqrt{8+2\, \sqrt{10+2\sqrt5}}+\sqrt{8-2\, \sqrt{10+2\sqrt5}}\; \; ,\; \; \; A>0\\\\\\A^2=\Big (\sqrt{8+2\, \sqrt{10+2\sqrt5}}+\sqrt{8-2\, \sqrt{10+2\sqrt5}}\Big )^2=\\\\=(8+2\sqrt{10+2\sqrt5})+(8-2\sqrt{10+2\sqrt5})+2\, \sqrt{8+2\, \sqrt{10+2\sqrt5}}\cdot \sqrt{8-2\, \sqrt{10+2\sqrt5}}=\\\\=16+2\, \sqrt{\, (\, 8+2\, \sqrt{10+2\sqrt5}\, )\, (\, 8-2\, \sqrt{10+2\sqrt5}\, )}=\\\\=16+2\, \sqrt{64-4(10+2\sqrt5)\, }\, =16+2\, \sqrt{24-8\sqrt5}\, =16+2\underbrace {\sqrt{4+20-2\cdot 2\cdot \sqrt{20}}}_{a^2+b^2-2ab}=$

$=16+2\, \sqrt{(2-\sqrt{20})^2}=16+2\cdot |2-\sqrt{20}|=\\\\=\Big [\; (2-\sqrt{20})<0\; \; \to \; \; \; |2-\sqrt{20}|=\sqrt{20}-2\; \Big ]=\\\\=16+2\cdot (\sqrt{20}-2)=12+2\sqrt{20}=2\cdot (6+\sqrt{20})=2\cdot (6+2\sqrt5)=\\\\=2\cdot (\underbrace {1+5+2\cdot 1\cdot \sqrt5}_{a^2+b^2+2ab})=2\cdot (1+\sqrt5)^2\\\\\\A^2=2\cdot (1+\sqrt5)^2\; \; ,\; \; A>0\; \; \Rightarrow \; \; \sqrt{A^2}=|A|=+A\\\\A=\sqrt{2\cdot (1+\sqrt5)^2}=\sqrt2\cdot \sqrt{(1+\sqrt5)^2}=\sqrt2\cdot |\underbrace {1+\sqrt5}_{>0}|=\sqrt2\cdot (1+\sqrt5)=\\\\=\sqrt2+\sqrt2\cdot \sqrt5=\sqrt2+\sqrt{10}$

$P.S.\; \; \; A=\underbrace {\sqrt{...}}_{>0}+\underbrace {\sqrt{..}}_{>0}>0\; \; ,\; \; tak\; kak\; \; \sqrt{...}\, >0\; .\\\\\\\sqrt{a}=|a|=\pm a=\left \{ {{+a\; ,\; esli\; a\geq 0\; ,} \atop {-a\; ,\; esli\; a<0\; .}} \right.$