Question

Colegio Naval
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Answer 1

Seja $p=\sqrt{10+2\sqrt{5}}.$ Do enunciado, segue que

$a=\sqrt{4-p}~~~~~~\mathbf{(i)}\\ \\ b=\sqrt{4+p}~~~~~~\mathbf{(ii)}$


Somando $\mathbf{(i)}$ e $\mathbf{(ii)},$ temos

$a+b=\sqrt{4-p}+\sqrt{4+p}\\ \\ (a+b)^{2}=\left(\sqrt{4-p}+\sqrt{4+p} \right )^{2}\\ \\ \\ (a+b)^{2}=\left(\sqrt{4-p} \right )^{2}+2\sqrt{4-p}\sqrt{4+p}+\left(\sqrt{4+p} \right )^{2}\\ \\ \\ (a+b)^{2}=4-p+2\sqrt{16-p^{2}}+4+p\\ \\ (a+b)^{2}=8+2\sqrt{16-p^{2}}\\\\ \\ (a+b)^{2}=8+2\sqrt{16-(10+2\sqrt{5})}\\ \\ \\ (a+b)^{2}=8+2\sqrt{6-2\sqrt{5}}\\ \\ \\ (a+b)^{2}=8+2q~~~~~~\mathbf{(iii)}$

onde $q=\sqrt{6-2\sqrt{5}}.$

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Da definição de $q,$ segue que

$q=\sqrt{6-2\sqrt{5}}\\ \\ q=\sqrt{5+1-2\sqrt{5}}\\ \\ q=\sqrt{(\sqrt{5})^{2}-2\sqrt{5}+1^{2}}\\ \\ q=\sqrt{(\sqrt{5}-1)^{2}}\\ \\ q=\sqrt{5}-1$


Portanto, voltando a $\mathbf{(iii)},$ temos que

$(a+b)^{2}=8+2\cdot (\sqrt{5}-1)\\ \\ (a+b)^{2}=8+2\sqrt{5}-2\\ \\ (a+b)^{2}=6+2\sqrt{5}\\ \\ (a+b)^{2}=5+1+2\sqrt{5}\\ \\ (a+b)^{2}=(\sqrt{5})^{2}+2\sqrt{5}+1^{2}\\ \\ (a+b)^{2}=(\sqrt{5}+1)^{2}$


Como $0<a<b\,,$ temos que

$a+b=\sqrt{5}+1$


Resposta: alternativa D.

Answer 2

$a = \sqrt{4 - \sqrt{10+2 \sqrt{5} } } \\ b= \sqrt{4+ \sqrt{10+2 \sqrt{5} } } \\ a+b = \sqrt{4- \sqrt{10+2 \sqrt{5} } } + \sqrt{4+ \sqrt{10+2 \sqrt{5} } } \\ (a+b)^2=[ \sqrt{4- \sqrt{10+2 \sqrt{5} } } + \sqrt{4+ \sqrt{10+2 \sqrt{5} } } ]^2 \\ (a+b)^2=4- \sqrt{10+2 \sqrt{5} } +2 \sqrt{16-10-2 \sqrt{5} } +4+ \sqrt{10+2 \sqrt{5} } \\ (a+b)^2=8+2 \sqrt{6-2 \sqrt{5} } \\ (a+b)^2=8+2\sqrt{5-2 \sqrt{5}+1 } \\ (a+b)^2=8+ 2\sqrt{ (\sqrt{5} -1)^2} \\ (a+b)^2=8+2( \sqrt{5}-1) \\ (a+b)^2=8+2 \sqrt{5} -2$
$(a+b)^2=6+2 \sqrt{5} \\ (a+b)^2=(5+2 \sqrt{5} +1) \\ (a+b)^2=( \sqrt{5} +1)^2 \\ a+b = \sqrt{5} +1$