Question

Exercitiul 3 multumesc❤️

Answer 1

 

$\displaystyle\\a=\Big(5-2\sqrt{6}\Big)^{1009}\cdot\Big(\sqrt{2}+\sqrt{3}\Big)^{2018}\cdot\Big(7-4\sqrt{3}\Big)^{100}\cdot\Big(\sqrt{3}+2\Big)^{200}\\\\a=\Big(5-2\sqrt{6}\Big)^{1009}\cdot\Big(\sqrt{2}+\sqrt{3}\Big)^{2\times1009}\cdot\Big(7-4\sqrt{3}\Big)^{100}\cdot\Big(\sqrt{3}+2\Big)^{2\times100}\\\\a=\Big(5-2\sqrt{6}\Big)^{1009}\cdot\left(\Big(\sqrt{2}+\sqrt{3}\Big)^2\right)^{1009}\cdot\Big(7-4\sqrt{3}\Big)^{100}\cdot\left(\Big(\sqrt{3}+2\Big)^2\right)^{100}$

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$\displaystyle\\a=\Big(5-2\sqrt{6}\Big)^{1009}\cdot\Big(2+2\sqrt{6}+3\Big)^{1009}\cdot\Big(7-4\sqrt{3}\Big)^{100}\cdot\Big(3+4\sqrt{3}+4\Big)^{100}\\\\a=\Big(5-2\sqrt{6}\Big)^{1009}\cdot\Big(5+2\sqrt{6}\Big)^{1009}\cdot\Big(7-4\sqrt{3}\Big)^{100}\cdot\Big(7+4\sqrt{3}\Big)^{100}\\\\a=\left[\Big(5-2\sqrt{6}\Big)\Big(5+2\sqrt{6}\Big)\right]^{1009}\cdot\left[\Big(7-4\sqrt{3}\Big)\Big(7+4\sqrt{3}\Big)\right]^{100}$

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$\displaystyle\\a=\left[\Big(5-2\sqrt{6}\Big)\Big(5+2\sqrt{6}\Big)\right]^{1009}\cdot\left[\Big(7-4\sqrt{3}\Big)\Big(7+4\sqrt{3}\Big)\right]^{100}\\\\a=\left[5^2-\Big(2\sqrt{6}\Big)^2\right]^{1009}\cdot\left[7^2-\Big(4\sqrt{3}\Big)^2\right]^{100}\\\\a=\left[25-24\right]^{1009}\cdot\left[49-48\right]^{100}\\\\a=1^{1009}\cdot1^{100}\\\\a=1\cdot1\\\\\\\boxed{\bf~a=1\in N}$

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