Question

(Colégio Naval) O valor de

$\dfrac{\big(3+2\sqrt{2}\big)^{2008}}{\big(5\sqrt{2}+7\big)^{\!1338}}+3-2\sqrt{2}$

(A) é múltiplo de 11.
(B) é múltiplo de 7.
(C) é múltiplo de 5.
(D) é múltiplo de 3.
(E) é primo.

Answer 1

Como é de praxe, eu sempre tento explicar a maioria (ou todos) dos passos realizados ao longo de cada uma das questões que resolvo. Todavia, para esta questão, não vejo necessidade alguma de explicar sequer um passo, isso porque cada um deles é, como veremos a seguir, demasiado trivial. Sem mais delongas, vamos à resolução:

$\tt\quad \dfrac{\big(3+2\sqrt{2}\:\!\big)^{\!2008}}{\big(5\sqrt{2}+7\big)^{\!1338}}+3-2\sqrt{2}\ \\\\\\\\ =\dfrac{\big(3+2\sqrt{2}\:\!\big)^{\!1338\:\!\:\!+\:\!\:\!670}}{\big(7+5\sqrt{2}\:\!\big)^{\!1338}}+3-2\sqrt{2}\\\\\\\\ =\dfrac{\big(3+2\sqrt{2}\:\!\big)^{\!1\!\:\!338}\cdot\, \big(3+2\sqrt{2}\:\!\big)^{\!670}}{\big(7+5\sqrt{2}\:\!\big)^{\!1338}}+3-2\sqrt{2}\\\\\\\\ =\dfrac{\big(3+2\sqrt{2}\:\!\big)^{\!1338}}{\big(7+5\sqrt{2}\big)^{\!1338}}\, \:\!\cdot\, \big(3+2\sqrt{2}\:\!\big)^{\!670}+3-2\sqrt{2}$

$\tt\!\!=\left(\dfrac{3+2\sqrt{2}}{7+5\sqrt{2}}\right)^{\!\!\!1338}\cdot\ \big(3+2\sqrt{2}\:\!\big)^{\!\:\!670}+3-2\sqrt{2}\\\\\\\\=\left(\dfrac{3+2\sqrt{2}}{7+5\sqrt{2}}\:\!\:\!\cdot\dfrac{7-5\sqrt{2}}{7-5\sqrt{2}}\right)^{\!\!\!1338}\cdot\ \big(3+2\sqrt{2}\;\!\big)^{\!670}+3-2\sqrt{2}\\\\\\\\=\bigg(\dfrac{21-15\sqrt{2}+14\sqrt{2}-20}{49-50}\bigg)^{\!\!\!1338}\cdot\ \big(3+2\sqrt{2}\:\!\big)^{\!670}+3-2\sqrt{2}\\\\\\\\=\big(\!\sqrt{2}-1\big)^{\!1338}\cdot\:\! \big(3+2\sqrt{2}\:\!\big)^{\!670}+3-2\sqrt{2}$

$= \tt\big(\!\sqrt{2}-1\big)^{\!1338}\cdot \,\big(2+2\sqrt{2}+1\big)^{\!670}+3-2\sqrt{2}\\\\\\ = \big(\!\sqrt{2}-1\big)^{\!1338}\cdot \:\! \left[\big(\!\:\!\sqrt{2}+1\big)^{\!\:\!2}\right]^{\!\:\!\:\!670}+3-2\sqrt{2}\\\\\\ =\left(\!\sqrt{2}-1\right)^{\!1338}\cdot\, \big(\!\:\!\sqrt{2}+1}\big)^{\!1340}+3-2\sqrt{2}\\\\\\ =\big(\!\sqrt{2}-1\big)^{\!1338}\cdot \, \big(\!\:\!\sqrt{2}+1\big)^{1338\:\!\:\!+\:\!2}+3-2\sqrt{2}$

$=\tt\big(\!\sqrt{2}-1\big)^{\!1338}\cdot\,\big(\!\:\!\sqrt{2}+1\big)^{\!1338}\cdot\, \big(\!\sqrt{2}+1\big)^{\!\:\!2}+3-2\sqrt{2}\\\\\\=\big[\big(\!\sqrt{2}-1\big)\cdot \big(\!\sqrt{2}+1\big)\big]^{\!1338}\cdot\,\big(2+2\sqrt{2}+1\big)+3-2\sqrt{2}\\\\\\ =(2-1)^{1338}\cdot \, \big(3+2\sqrt{2}\:\!\big)+3-2\sqrt{2}\\\\\\=1^{\!1338}\cdot \, \big(3+2\sqrt{2}\:\!\big)+3-2\sqrt{2}\\\\\\=1\cdot\big(3+2\sqrt{2}\:\!\big)+3-2\sqrt{2}\\\\\\ =3+2\sqrt{2}+3-2\sqrt{2}\\\\\\=6\ \rightarrow\ \'e\ m\'ultiplo\ de\ 3.$

Resposta: letra D).