Exercițiul 1:
a)
[tex]1.02 + \dfrac{49}{50} - 1.1(6) \times \dfrac{3}{7} = \\ \\ \\ \dfrac{102}{100}^{(2} + \dfrac{49}{50} - \dfrac{116 - 11}{90} \times \dfrac{3}{7} = \\ \\ \\ \dfrac{51}{50} + \dfrac{49}{50} - \dfrac{105}{90} ^{(5} \times \dfrac{3}{7} = \\ \\ \\ \dfrac{100}{50} ^{(50} - \dfrac{21}{18} ^{(3} \times \dfrac{3}{7} = \\ \\ \\ ^{6)} 2 - \dfrac{7}{6} \times \dfrac{3}{7} = \\ \\ \\ \dfrac{12}{6} - \dfrac{7}{6} \times \dfrac{3}{7} = \\ \\ \\ \dfrac{5}{6} \times \dfrac{3}{7} = \dfrac{15}{42} ^{(3} = \dfrac{5}{14} [/tex]
b)
[tex]( - \dfrac{3}{7} ) ^{5} \times (( - \dfrac{3}{7} )^{2} )^{4} \div ( - \dfrac{3}{7} )^{12} = \\ \\ \\ ( - \dfrac{3}{7} ) ^{5} \times ( - \dfrac{3}{7} )^{2 \times 4} \div ( - \dfrac{3}{7} )^{12} = \\ \\ \\ ( - \dfrac{3}{7} )^{5} \times ( - \dfrac{3}{7} )^{8} \div ( - \dfrac{3}{7} ) ^{12} = \\ \\ \\ ( - \dfrac{3}{7} )^{5 + 8} \div ( - \dfrac{3}{7} ) ^{12} = \\ \\ \\ ( - \dfrac{3}{7} )^{13} \div ( - \dfrac{3}{7} ) ^{12} = \dfrac{3}{7} [/tex]
Exercițiul 2 :
[tex]( ^{2)} 1 - \dfrac{1}{2} ) \times ( ^{3)} 1 - \dfrac{1}{3} ) \times (^{4)} 1 - \dfrac{1}{4} ) \times ( ^{5)} 1 - \dfrac{1}{5} ) > \dfrac{1}{6} \\ \\ \\ ( \dfrac{2}{2} - \dfrac{1}{2} ) \times ( \dfrac{3}{3} - \dfrac{1}{3} ) \times ( \dfrac{4}{4} - \dfrac{1}{4} ) \times ( \dfrac{5}{5} - \frac{1}{5} ) > \dfrac{1}{6} \\ \\ \\ \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} > \dfrac{1}{6} \\ \\ \\ ^{6)} \dfrac{1}{5} > ^{5)} \dfrac{1}{6} \\ \\ \\ \dfrac{6}{30} > \dfrac{5}{30} [/tex]
Exercițiul 3:
[tex]a = - 0.1 \times \dfrac{25}{2} \\ \\ \\ a = - \dfrac{1}{10} \times \dfrac{25}{2} \\ \\ \\ a = - \dfrac{5}{4} [/tex]
[tex]b = ( - \dfrac{25}{2} )^{ - 2} \\ \\ \\ b = \dfrac{4}{625} [/tex]
[tex]a < b...a = negativ..b = pozitiv[/tex]
