363* Вправа 363Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 99

363. Обчисліть:

1) ${\displaystyle \frac{8^{-4}\cdot 6^{10}+{27}^3}{{10}^0\cdot {81}^3-{18}^6\cdot 6^{-3}}\cdot {\left(\frac{7}{19}\right)}^{-1}}$;

чисельник: ${\displaystyle 8^{-4}\cdot 6^{10}=}$${\displaystyle 2^{-12}\cdot 2^{10}\cdot 3^{10}=}$${\displaystyle \frac{3^{10}}{4}}$; ${\displaystyle {27}^3=}$${\displaystyle 3^9}$;

${\displaystyle \frac{3^{10}}{4}+3^9=}$${\displaystyle 3^9(\frac{3}{4}+1)=}$${\displaystyle 3^9\cdot \frac{7}{4}}$;

знаменник: ${\displaystyle {81}^3=}$${\displaystyle 3^{12}}$; ${\displaystyle {18}^6\cdot 6^{-3}=}$${\displaystyle 2^6\cdot 3^{12}\cdot 2^{-3}\cdot 3^{-3}=}$${\displaystyle 8\cdot 3^9}$;

${\displaystyle 3^{12}-8\cdot 3^9=}$${\displaystyle 3^9(27-8)=}$${\displaystyle 19\cdot 3^9}$;

${\displaystyle \frac{\cancel{3^9}\cdot \stackrel{1}{\cancel{7}}}{4\cdot \underset{1}{\cancel{19}}\cdot \cancel{3^9}}\cdot \frac{\stackrel{1}{\cancel{19}}}{\underset{1}{\cancel{7}}}=}$${\displaystyle \frac{1}{4}=}$${\displaystyle 0,25}$;

2) ${\displaystyle \frac{4^{-8}\cdot {0,0001}^{-5}+{125}^6\cdot {\left(3\frac{1}{3}\right)}^2}{{\left({0,2}^{10}\right)}^{-2}+3^{-2}\cdot {50}^{10}\cdot {0,5}^{12}}}$;

чисельник: ${\displaystyle 4^{-8}\cdot {0,0001}^{-5}=}$${\displaystyle 2^{-16}\cdot {10}^{20}=}$${\displaystyle 2^4\cdot 5^{20}}$;

${\displaystyle {125}^6\cdot {\left(\frac{10}{3}\right)}^2=}$${\displaystyle 5^{18}\cdot \frac{2^2\cdot 5^2}{9}=}$${\displaystyle \frac{4\cdot 5^{20}}{9}}$;

${\displaystyle 16\cdot 5^{20}+\frac{4\cdot 5^{20}}{9}=}$${\displaystyle 5^{20}\cdot \frac{148}{9}}$;

знаменник: ${\displaystyle {\left({0,2}^{10}\right)}^{-2}=}$${\displaystyle {0,2}^{-20}=}$${\displaystyle 5^{20}}$; ${\displaystyle 3^{-2}\cdot {50}^{10}\cdot {0,5}^{12}=}$${\displaystyle \frac{2^{10}\cdot 5^{20}}{9\cdot 2^{12}}=}$${\displaystyle \frac{5^{20}}{36}}$;

${\displaystyle 5^{20}+\frac{5^{20}}{36}=}$${\displaystyle 5^{20}\cdot \frac{37}{36}}$;

${\displaystyle \frac{148}{9}:\frac{37}{36}=}$${\displaystyle \frac{148\cdot 36}{9\cdot 37}=}$${\displaystyle \frac{4\cdot \stackrel{4}{\cancel{148}}\cdot \stackrel{1}{\cancel{9}}}{\underset{1}{\cancel{9}}\cdot \underset{1}{\cancel{37}}}=}$${\displaystyle 16}$.