Explicație pas cu pas:
a)
[tex]a = \sqrt{6} \cdot\left( \frac{1}{ \sqrt{3} } + \frac{1}{ \sqrt{2} }\right) - \sqrt{3} \cdot \left( \frac{1}{ \sqrt{2}} + \frac{2}{ \sqrt{6}} \right) - \sqrt{2}\cdot\left( \frac{1}{ \sqrt{6}} - \frac{3}{2 \sqrt{3} }\right) \\ = \frac{ \sqrt{6} }{ \sqrt{3} } + \frac{ \sqrt{6} }{ \sqrt{2} } - \frac{ \sqrt{3}}{ \sqrt{2} } - \frac{2 \sqrt{3} }{ \sqrt{6} } - \frac{ \sqrt{2} }{ \sqrt{6} } + \frac{3 \sqrt{2} }{2 \sqrt{3} } \\ = \frac{2 \sqrt{2}\cdot \sqrt{6} + 2 \sqrt{3}\cdot \sqrt{6} - 2 \sqrt{3}\cdot \sqrt{3} - 2\cdot 2\sqrt{3} - 2\cdot \sqrt{2} + \sqrt{2}\cdot3 \sqrt{2} }{2 \sqrt{6} } \\ = \frac{4 \sqrt{3} + 6 \sqrt{2} - 6 - 4 \sqrt{3} - 2 \sqrt{2} + 6}{2 \sqrt{6} } = \frac{4 \sqrt{2} }{2 \sqrt{6} } = \frac{2}{ \sqrt{3} } [/tex]
[tex] \Rightarrow a\in \mathbb{R - Q}[/tex]
b)
[tex]b = | \left( - \frac{5}{6} \right)^{ - 1} | = | - \frac{6}{5}| = \frac{6}{5} \\ [/tex]
[tex]{a}^{2} = \left(\frac{2}{ \sqrt{3} } \right)^{2} = \frac{4}{3} \\ {b}^{2} = \frac{36}{25} \\ \frac{4\cdot25}{3\cdot25} = \frac{100}{75} < \frac{36\cdot3}{25\cdot3} = \frac{108}{75} \\ \Rightarrow a < b[/tex]
