Verified answer

Explicație pas cu pas:

a)

[tex]a = \sqrt{ {( - \sqrt{3} )}^{2} } - {\Big( - \dfrac{2}{3}\Big)}^{ - 2} + \dfrac{9}{4} - 2 \sqrt{3} = \\ [/tex]

[tex]= \sqrt{3} - {\Big( - \dfrac{3}{2}\Big)}^{2} + \dfrac{9}{4} - 2 \sqrt{3} [/tex]

[tex]= \sqrt{3} - \dfrac{9}{4} + \dfrac{9}{4} - 2 \sqrt{3} = \bf - \sqrt{3} [/tex]

[tex]{a}^{2} = {\Big( - \sqrt{3} \Big)}^{2} = 3[/tex]

[tex]{a}^{3} = {a}^{2} \cdot a = 3 \cdot ( - \sqrt{3}) = - 3 \sqrt{3} [/tex]

[tex]{a}^{4} = {{a}^{2}}^{2} = {3}^{2} = 9[/tex]

[tex]{a}^{5} = {a}^{4} \cdot a = 9 \cdot ( - \sqrt{3}) = - 9 \sqrt{3} [/tex]

[tex]{a}^{6} = {a}^{4} \cdot {a}^{2} = 9 \cdot 3 = 27[/tex]

[tex]{a}^{7} = {a}^{6} \cdot a = 27 \cdot ( - \sqrt{3}) = - 27 \sqrt{3} [/tex]

[tex]{a}^{8} = {{a}^{4}}^{2} = {9}^{2} = 81[/tex]

[tex]{a}^{9} = {a}^{8} \cdot a = 81 \cdot ( - \sqrt{3}) = - 81 \sqrt{3}[/tex]

[tex]{a}^{10} = {a}^{8} \cdot {a}^{2} = 81 \cdot 3 = 243[/tex]

b)

[tex]{a}^{ - 2} + {a}^{ - 4} + {a}^{ - 6} + {a}^{ - 8} = {( - \sqrt{3})}^{ - 2} + {( - \sqrt{3})}^{ - 4} + {( - \sqrt{3})}^{ - 6} + {( - \sqrt{3})}^{ - 8} = {(\sqrt{3})}^{ - 2} + {(\sqrt{3})}^{ - 4} + {(\sqrt{3})}^{ - 6} + {(\sqrt{3})}^{ - 8} = {\Big(\dfrac{1}{ \sqrt{3} }\Big)}^{2} + {\Big(\dfrac{1}{ \sqrt{3} }\Big)}^{4} + {\Big(\dfrac{1}{ \sqrt{3} }\Big)}^{6} + {\Big(\dfrac{1}{ \sqrt{3} }\Big)}^{8} = \dfrac{1}{({\sqrt{3})}^{2} } + \dfrac{1}{({\sqrt{3})}^{4} } + \dfrac{1}{({\sqrt{3})}^{6} } + \dfrac{1}{({\sqrt{3})}^{8} } = \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} \in \mathbb{Q} [/tex]

q.e.d.