Алгебра. 67 баллов. Объясните.

Question 1

Question 2

Решите задачу:

$a) \: 2 \frac{2}{3} {x}^{5} {y}^{8} {(- 1 \frac{1}{2} {x}^{3} {y}^{6} )}^{2} = 6 {x}^{11} {y}^{20} \\ \frac{8}{3} {x}^{5} {y}^{8}( { - \frac{3}{2} )}^{2} {({x}^{3})}^{2} {({y}^{6})}^{2} = 6 {x}^{11} {y}^{20} \\ \frac{8}{3} {x}^{5} {y}^{8} \times \frac{9}{4} {x}^{6} {y}^{12} = 6 {x}^{11} {y}^{20} \\ \frac{8}{3} \times \frac{9}{4} {x}^{5 + 6} {y}^{8 + 12} = 6 {x}^{11} {y}^{20} \\ 6 {x}^{11} {y}^{20} = 6 {x}^{11} {y}^{20}$

$б) \: - 3 \frac{1}{5} {a}^{8} n \times ( { \frac{1}{2} {a}^{3} {n}^{8}) }^{4} = - \frac{1}{5} {a}^{20} {n}^{33} \\ - \frac{16}{5} {a}^{8} n \times( { \frac{1}{2}) }^{4}( { {a}^{3}) }^{4} ( { {n}^{8}) }^{4} = - \frac{1}{5} {a}^{20} {n}^{33} \\ - \frac{16}{5} {a}^{8} n \times \frac{1}{16} {a}^{12} {n}^{32} = - \frac{1}{5} {a}^{20} {n}^{33} \\ - \frac{16}{5} \times \frac{1}{16} {a}^{8 + 12} {n}^{1 + 32} = - \frac{1}{5} {a}^{20} {n}^{33} \\ - \frac{1}{5} {a}^{20} {n}^{33} = - \frac{1}{5} {a}^{20} {n}^{33}$

$в) \: ( {2 \frac{1}{3} {a}^{4} {c}^{8}) }^{2} ( - 1 \frac{2}{7} {a}^{5} {c}^{12}) = - 7{a}^{13} {c}^{28} \\ ( { \frac{7}{3} )}^{2} {({a}^{4})}^{2} ( {{c}^{8})}^{2} ( - \frac{9}{7} {a}^{5} {c}^{12}) = - 7{a}^{13} {c}^{28} \\ \frac{49}{9} {a}^{8} {c}^{16} \times ( - \frac{9}{7} {a}^{5} {c}^{12})= - 7{a}^{13} {c}^{28} \\ \frac{49}{9} \times ( - \frac{9}{7}) {a}^{8 + 5} {c}^{16 + 12} = - 7{a}^{13} {c}^{28} \\ - 7{a}^{13} {c}^{28} = - 7{a}^{13} {c}^{28}$