Question

va rog, putin ajutor? ​

Answer 1

 

$\displaystyle\bf\\a=\left(5-2\sqrt{6}\right)^{1009}\cdot\left(\sqrt{2}+\sqrt{3}\right)^{2018}\cdot\left(7-4\sqrt{3}\right)^{100}\cdot\left(\sqrt{3}+2\right)^{200}\\\\\\a=\left(3+2-2\sqrt{3}\sqrt{2}\right)^{1009}\cdot\left(\sqrt{3}+\sqrt{2}\right)^{2018}\cdot\left(4+3-2\cdot2\sqrt{3}\right)^{100}\cdot\left(2+\sqrt{3}\right)^{200}\\\\\\a=\left(3-2\sqrt{3}\sqrt{2}+2\right)^{1009}\cdot\left(\sqrt{3}+\sqrt{2}\right)^{2018}\cdot\left(4-2\cdot2\sqrt{3}+3\right)^{100}\cdot\left(2+\sqrt{3}\right)^{200}$

 

$\displaystyle\bf\\a=\left(\Big(\sqrt{3}-\sqrt{2}\Big)^2\right)^{1009}\cdot\left(\sqrt{3}+\sqrt{2}\right)^{2018}\cdot\left(\Big(2-\sqrt{3}\Big)^2\right)^{100}\cdot\left(2+\sqrt{3}\right)^{200}\\\\\\a=\left(\sqrt{3}-\sqrt{2}\right)^{2018}\cdot\left(\sqrt{3}+\sqrt{2}\right)^{2018}\cdot\left(2-\sqrt{3}\right)^{200}\cdot\left(2+\sqrt{3}\right)^{200}$

 

$\displaystyle\bf\\a=\left[\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)\right]^{2018}\cdot\left[\left(2-\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\right]^{200}\\\\\\a=\left[\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\right]^{2018}\cdot\left[2^2-\left(\sqrt{3}\right)^2\right]^{200}\\\\\\a=\Big[3-2\Big]^{2018}\cdot\Big[4-3\Big]^{200}\\\\\\a=1^{2018}\cdot1^{200}\\\\a=1\cdot1\\\\\boxed{\bf~a=1\in N}$