Question

ajutor urgent
4. Arătaţi că numărul n = √5-2√6 +√√5+2√6-√(3 +2√3)² este intreg.​

Answer 1

$\displaystyle n=\sqrt{5-2\sqrt{6} }+\sqrt{5+2\sqrt{6} } -\sqrt{\Big(3+2\sqrt{3} \Big)^2} \\\\ n=\sqrt{\frac{5+\sqrt{5^2-2^2 \cdot 6} }{2} } -\sqrt{\frac{5-\sqrt{5^2-2^2 \cdot 6} }{2} }+\sqrt{\frac{5+\sqrt{5^2-2^2 \cdot 6} }{2} }+\\\\ +\sqrt{\frac{5-\sqrt{5^2-2^2 \cdot 6} }{2} }-\Big|3+2\sqrt{3} \Big|\\\\ n=\sqrt{\frac{5+\sqrt{25-4 \cdot 6} }{2} } -\sqrt{\frac{5-\sqrt{25-4 \cdot 6} }{2} } +\sqrt{\frac{5+\sqrt{25-4 \cdot 6} }{2} } +\\\\ +\sqrt{\frac{5-\sqrt{25-4 \cdot 6} }{2} } -\Big(3+2\sqrt{3} \Big)$

$\displaystyle n=\sqrt{\frac{5+\sqrt{25-24} }{2} } -\sqrt{\frac{5-\sqrt{25-24} }{2} } +\sqrt{\frac{5+\sqrt{25-24} }{2} } +\\\\+\sqrt{\frac{5-\sqrt{25-24} }{2} } -3-2\sqrt{3} \\\\ n=\sqrt{\frac{5+\sqrt{1} }{2} }-\sqrt{\frac{5-\sqrt{1} }{2} }+\sqrt{\frac{5+\sqrt{1} }{2} }+\sqrt{\frac{5-\sqrt{1} }{2} }-3-2\sqrt{3} \\\\n=\sqrt{\frac{5+1}{2} }- \sqrt{\frac{5-1}{2} }+\sqrt{\frac{5+1}{2} }+\sqrt{\frac{5-1}{2} }-3-2\sqrt{3}$

$\displaystyle n= \sqrt{\frac{6}{2} } -\sqrt{\frac{4}{2} } +\sqrt{\frac{6}{2} } +\sqrt{\frac{4}{2} } -3-2\sqrt{3} \\\\n=\sqrt{3} -\sqrt{2} +\sqrt{3} +\sqrt{2} -3-2\sqrt{3} \\\\ n=-3\in\mathbb{Z}$