359 Вправа 359Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 98

359. Знайдіть x. Подайте відповідь у вигляді степеня.

1) ${\displaystyle {({0,1}^{-8}:{25}^4\cdot {12,5}^0)}^{-1}=}$${\displaystyle x\cdot 2^5}$;

${\displaystyle {0,1}^{-8}:{25}^4\cdot 1=}$${\displaystyle {10}^8:5^8=}$${\displaystyle {(10:5)}^8=}$${\displaystyle 2^8}$;

${\displaystyle {\left(2^8\right)}^{-1}=}$${\displaystyle x\cdot 2^5}$ | ${\displaystyle :2^5}$

${\displaystyle x=}$${\displaystyle 2^{-8}:2^5=}$${\displaystyle 2^{-13}}$;

2) ${\displaystyle {63}^{-12}=}$${\displaystyle {21}^{-12}\cdot x}$ | ${\displaystyle :{21}^{-12}}$

${\displaystyle x=}$${\displaystyle {\left(\frac{63}{21}\right)}^{-12}=}$${\displaystyle 3^{-12}}$;

3) ${\displaystyle {(-30)}^{-2}=}$${\displaystyle {25}^{-1}\cdot x}$ | ${\displaystyle :{25}^{-1}}$

${\displaystyle x=}$${\displaystyle \frac{25}{{30}^2}=}$${\displaystyle \frac{25}{900}=}$${\displaystyle \frac{1}{36}=}$${\displaystyle 6^{-2}}$;

4) ${\displaystyle {10}^{-28}=}$${\displaystyle 5^{-28}\cdot x}$ | ${\displaystyle :5^{-28}}$

${\displaystyle x=}$${\displaystyle {\left(\frac{10}{5}\right)}^{-28}=}$${\displaystyle 2^{-28}}$;

5) ${\displaystyle {15}^{-6}=}$${\displaystyle x\cdot 3^{-6}}$ | ${\displaystyle :3^{-6}}$

${\displaystyle x=}$${\displaystyle {\left(\frac{15}{3}\right)}^{-6}=}$${\displaystyle 5^{-6}}$;

6) ${\displaystyle {\left({0,6}^{-5}\right)}^3=}$${\displaystyle {\left(\frac{1}{5^{-3}}\right)}^5\cdot x}$; ${\displaystyle {0,6}^{-15}=}$${\displaystyle 5^{15}\cdot x}$ | ${\displaystyle :5^{15}}$

${\displaystyle x=}$${\displaystyle {0,6}^{-15}\cdot 5^{-15}=}$${\displaystyle {(0,6\cdot 5)}^{-15}=}$${\displaystyle 3^{-15}}$;

7) ${\displaystyle {(0,4\cdot 3)}^{-8}\cdot {0,05}^8=}$${\displaystyle x\cdot 6^{-8}}$;

${\displaystyle {1,2}^{-8}\cdot {0,05}^8=}$${\displaystyle {\left(\frac{0,05}{1,2}\right)}^8=}$${\displaystyle {\left(\frac{1}{24}\right)}^8=}$${\displaystyle {24}^{-8}}$;

${\displaystyle {24}^{-8}=}$${\displaystyle x\cdot 6^{-8}}$ | ${\displaystyle :6^{-8}}$

${\displaystyle x=}$${\displaystyle {\left(\frac{24}{6}\right)}^{-8}=}$${\displaystyle 4^{-8}}$;

8) ${\displaystyle {24}^{-13}=}$${\displaystyle x\cdot 2^{-26}\cdot 3^{-13}}$; ${\displaystyle 2^{-26}\cdot 3^{-13}=}$${\displaystyle {(2^2\cdot 3)}^{-13}=}$${\displaystyle {12}^{-13}}$;

${\displaystyle x=}$${\displaystyle {\left(\frac{24}{12}\right)}^{-13}=}$${\displaystyle 2^{-13}}$;

9) ${\displaystyle {64}^{-4}\cdot 4^{12}\cdot {16}^{-2}\cdot x=}$${\displaystyle 4^{-12}\cdot {16}^5}$; в основі 2:

${\displaystyle 2^{-24}\cdot 2^{24}\cdot 2^{-8}\cdot x=}$${\displaystyle 2^{-24}\cdot 2^{20}}$;

${\displaystyle 2^{-8}\cdot x=}$${\displaystyle 2^{-4}}$ | ${\displaystyle :2^{-8}}$

${\displaystyle x=}$${\displaystyle 2^{-4+8}=}$${\displaystyle 2^4}$;

10) ${\displaystyle x=}$${\displaystyle \frac{{27}^{-5}\cdot 9^3}{{81}^{-2}}=}$${\displaystyle \frac{3^{-15}\cdot 3^6}{3^{-8}}=}$${\displaystyle 3^{-15+6+8}=}$${\displaystyle 3^{-1}}$;

11) ${\displaystyle {(0,25x)}^{10}=}$${\displaystyle {\left(\frac{4}{3}\right)}^{-10}=}$${\displaystyle {\left(\frac{3}{4}\right)}^{10}}$;

${\displaystyle 0,25x=}$${\displaystyle \frac{3}{4}}$ або ${\displaystyle 0,25x=}$${\displaystyle -\frac{3}{4}}$ | ${\displaystyle :0,25}$

${\displaystyle x=}$${\displaystyle 3}$ або ${\displaystyle x=}$${\displaystyle -3}$;

12) ${\displaystyle {\left(\frac{x}{7}\right)}^8=}$${\displaystyle {\left(\frac{25}{49}\right)}^8}$;

${\displaystyle \frac{x}{7}=}$${\displaystyle \frac{25}{49}}$ або ${\displaystyle \frac{x}{7}=}$${\displaystyle -\frac{25}{49}}$ | ${\displaystyle \cdot 7}$

${\displaystyle x=}$${\displaystyle \frac{25}{7}=}$${\displaystyle 5^2\cdot 7^{-1}}$ або ${\displaystyle x=}$${\displaystyle -5^2\cdot 7^{-1}}$;

13) ${\displaystyle 5^4=}$${\displaystyle 5x^3}$ | ${\displaystyle :5}$

${\displaystyle x^3=}$${\displaystyle 5^3}$; ${\displaystyle x=}$${\displaystyle 5}$;

14) ${\displaystyle {\left(1\frac{1}{3}\right)}^{-3}=}$${\displaystyle {\left(\frac{3}{4}\right)}^3=}$${\displaystyle \frac{27}{64}}$; ${\displaystyle {0,75}^3=}$${\displaystyle {\left(\frac{3}{4}\right)}^3=}$${\displaystyle \frac{27}{64}}$;

${\displaystyle x=}$${\displaystyle \frac{27}{64}-\frac{27}{64}=}$${\displaystyle 0}$.