Question

Ajutor cu aceste calcule! Mulțumesc!

Answer 1

$a = {}^{ {}^{1 - \sqrt{2}) } } \tfrac{2 - \sqrt{2} }{1 + \sqrt{2} } + {}^{ {}^{ \sqrt{2}) } } \tfrac{2( \sqrt{2} + 2)}{ \sqrt{2} } - {}^{ {}^{ \sqrt{2}) } } \tfrac{16 - 3 \sqrt{2} }{2 \sqrt{2} } \\ \\ a = \tfrac{(2 - \sqrt{2} )(1 - \sqrt{2} }{1 {}^{2} - \sqrt{2} {}^{2} } + \tfrac{(2 \sqrt{2} + 4) \sqrt{2} }{2} - \tfrac{(16 - 3 \sqrt{2} ) \sqrt{2} }{4} \\ \\ a \small= - (2 - 2 \sqrt{2} - \sqrt{2} + 2) + ( \sqrt{2} + 2) \sqrt{2} \\ - \tfrac{16 \sqrt{2} - 6}{4} \\ \\ a = \small - 4 + 3 \sqrt{2} + 2 + 2 \sqrt{2} - \tfrac{8 \sqrt{2} - 3}{2} \\ \\ a = \small - 2 + 5 \sqrt{2} - \tfrac{8 \sqrt{2} - 3 }{2} \\ \\ a = \tfrac{ - 4 + 10 \sqrt{2} - (8 \sqrt{2} - 3) }{2} \\ \\ a = \tfrac{ - 4 + 10 \sqrt{2} - 8 \sqrt{2} + 3 }{2} \\ \\ a = \tfrac{ - 1 + 2 \sqrt{2} }{2} \\ \\\boxed{ a = \tfrac{2 \sqrt{2} - 1 }{2}} \\ \\ \\ \\ \\ \\ b = {}^{ {}^{ \sqrt{2}) } } \tfrac{4 - \sqrt{2} }{2 \sqrt{2} } + {}^{ {}^{2 - \sqrt{2} )} } \tfrac{2 - \sqrt{2} }{2 + \sqrt{2} } \\ \\ b = \tfrac{4 \sqrt{2} - 2}{4} + \tfrac{(2 - \sqrt{2} ) {}^{2} }{2 {}^{2} - 2} \\ \\ b = \tfrac{2 \sqrt{2} - 1 }{2} + \tfrac{4 - 4 \sqrt{2} + 2 }{2} \\ \\ b = \tfrac{2 \sqrt{2} - 1 + 6 - 4 \sqrt{2} }{2} \\ \\ b = \tfrac{ - 2 \sqrt{2} + 5 }{2} \\ \\ \\ \implies \: n = \: a + b \\ \\ \implies \: n = \tfrac{2 \sqrt{2} - 1 }{2} + \tfrac{ - 2 \sqrt{2} + 5 }{2} \\ \\ \implies \: n = \tfrac{2 \sqrt{2} - 1 - 2 \sqrt{2} + 5 }{2} \\ \\ \implies \: n = \tfrac{6}{2} \\ \\ \implies \:\boxed{ n = 3\in\mathbb{N}}$