Răspuns:
2
Explicație pas cu pas:
[tex]a = {\Big( \frac{2}{3} \Big)}^{100} : \frac{ {2}^{101} + {6}^{101} }{{3}^{101} + {9}^{101}} = {\Big( \frac{2}{3} \Big)}^{100} \cdot \frac{ {3}^{101} + {(3 \cdot 3)}^{101} }{{2}^{101} + {(2 \cdot 3)}^{101}} = \\ = {\Big( \frac{2}{3} \Big)}^{100} \cdot \frac{ {3}^{101} \cdot (1 + {3}^{101})}{{2}^{101} \cdot (1 + {3}^{101})} = \frac{ {2}^{100} }{ {3}^{100} } \cdot \frac{ {3}^{101} }{ {2}^{101} } = \bf \frac{3}{2} [/tex]
[tex]3a - {a}^{2} - \frac{1}{4} = 3 \cdot \frac{3}{2} - {\Big( \frac{3}{2} \Big)}^{2} - \frac{1}{4} = \\ = \frac{^{2)} 9}{2} - \frac{9}{4} - \frac{1}{4} = \frac{18 - 10}{4} = \frac{8}{4} = \bf 2[/tex]
