Question

$x=\sqrt{\sqrt{11} -4\sqrt{6+2\sqrt{4-2\sqrt{3} } } } \\y=\sqrt{\sqrt{11} +4\sqrt{6-2\sqrt{4+2\sqrt{3} } } }$


a) x∈R\Q, y∈R\Q

b) x+y∈Q si xy∈Q

REPEDE, DAU CORONITA!!!

Answer 1

$a)\\x = \sqrt{11-4\sqrt{6+2\sqrt{4-2\sqrt 3}}} \\ \\ \sqrt{4-2\sqrt{3}} = \sqrt{(1-\sqrt{3})^2} = |1-\sqrt{3}| = \sqrt{3}-1 \\ \sqrt{6+2(\sqrt{3}-1)} = \sqrt{2\sqrt{3}+4} = \sqrt{(1+\sqrt{3})^2} = 1+\sqrt{3}\\\\ \Rightarrow x = \sqrt{11-4(1+\sqrt{3})} = \sqrt{7-4\sqrt{3}} = \sqrt{(2-\sqrt{3})^2}= |2-\sqrt{3}| \\ \\ \Rightarrow x = 2-\sqrt{3}\in \mathbb{R}\backslash\mathbb{Q}$

$\\y = \sqrt{11+4\sqrt{6-2\sqrt{4+2\sqrt 3}}} \\ \\ \sqrt{4+2\sqrt{3}} = \sqrt{(1+\sqrt{3})^2} = 1+\sqrt{3} \\ \sqrt{6-2(1+\sqrt{3})} = \sqrt{4-2\sqrt{3}} = \sqrt{(1-\sqrt{3})^2} = |1-\sqrt{3}|=\sqrt{3}-1\\\\ \Rightarrow y = \sqrt{11+4(\sqrt{3}-1)} = \sqrt{7+4\sqrt{3}} = \sqrt{(2+\sqrt{3})^2}= 2+\sqrt{3} \\ \\ \Rightarrow y = 2+\sqrt{3}\in \mathbb{R}\backslash\mathbb{Q}$

$\\b)\\\\x+y = (2-\sqrt{3})+(2+\sqrt{3}) = 2+2-\sqrt{3}+\sqrt{3} = 4\in \mathbb{Q}\\\\xy = (2-\sqrt{3})(2+\sqrt{3}) = 2^2-(\sqrt{3})^2 = 4-3 = 1 \in \mathbb{Q}$