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Álvaro Antônio de Lacerda Rosário 2025-01-06 19:32:20 -03:00
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Matemática/1. Tópicos Fundamentais/Exercícios/Aritmética e Álgebra.md
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@ -30,12 +30,15 @@ a_1 + a_2 &= \dfrac{1}{\sqrt{b}+\sqrt{c}} + \dfrac{1}{\sqrt{a}+\sqrt{b}} \\
&= \dfrac{(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})} + \dfrac{(\sqrt{b}+\sqrt{c})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})} \\ &= \dfrac{(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})} + \dfrac{(\sqrt{b}+\sqrt{c})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})} \\
&= \dfrac{\sqrt{a}+ 2\sqrt{b}+\sqrt{c}}{(\sqrt{a}\cdot+\sqrt{b})(\sqrt{b}+\sqrt{c})} &= \dfrac{\sqrt{a}+ 2\sqrt{b}+\sqrt{c}}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})}
\end{align*} \end{align*}
$$ $$
$$ $$
= \dfrac{\sqrt{a}+ 2\sqrt{b}+\sqrt{c}}{(\sqrt{a}\cdot+\sqrt{b})(\sqrt{b}+\sqrt{c})} \dfrac{2}{\sqrt{c} + \sqrt{a}} = \dfrac{\sqrt{a}+ 2\sqrt{b}+\sqrt{c}}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})}
$$ $$
$$
2(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c}) = (\sqrt{a}+ 2\sqrt{b}+\sqrt{c})(\sqrt{c} + \sqrt{a})
$$
$$
2(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c}) = (\sqrt{a}+ 2\sqrt{b}+\sqrt{c})(\sqrt{c} + \sqrt{a})
$$