365* Вправа 365Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 99

365. Спростіть вираз:

Спрощуємо від внутрішніх дужок до зовнішніх.

1) ${\displaystyle {({(a^{-1}\cdot a)}^{-1}\cdot a)}^{-1}\cdot a=}$${\displaystyle {({\left(a^0\right)}^{-1}\cdot a)}^{-1}\cdot a=}$${\displaystyle a^{-1}\cdot a=}$${\displaystyle 1}$;

2) ${\displaystyle {({(a^{-2}\cdot a^2)}^{-5}\cdot a^2)}^{-1}=}$${\displaystyle {({\left(a^0\right)}^{-5}\cdot a^2)}^{-1}=}$${\displaystyle {\left(a^2\right)}^{-1}=}$${\displaystyle a^{-2}}$;

3) ${\displaystyle {({(x^3\cdot x^{-7})}^{-5}\cdot x^{-1})}^{10}=}$${\displaystyle {({\left(x^{-4}\right)}^{-5}\cdot x^{-1})}^{10}=}$${\displaystyle {(x^{20}\cdot x^{-1})}^{10}=}$${\displaystyle {\left(x^{19}\right)}^{10}=}$${\displaystyle x^{190}}$;

4) ${\displaystyle ({(x^{-9}:x^{-1})}^{-1}:x^7):x^{10}=}$${\displaystyle ({\left(x^{-8}\right)}^{-1}:x^7):x^{10}=}$${\displaystyle x^8:x^7:x^{10}=}$${\displaystyle x^{8-7-10}=}$${\displaystyle x^{-9}}$;

5) ${\displaystyle {({({\left(a^{-3}\right)}^{-1}\cdot b^3)}^{-4}\cdot a^{12})}^2\cdot b^{20}=}$${\displaystyle {({\left(a^3{b}^3\right)}^{-4}\cdot a^{12})}^2\cdot b^{20}=}$${\displaystyle {\left(a^{-12}{b}^{-12}{a}^{12}\right)}^2\cdot b^{20}=}$${\displaystyle b^{-24}\cdot b^{20}=}$${\displaystyle b^{-4}}$;

6) ${\displaystyle a\cdot {\left(\frac{a}{b}\right)}^{-2}=}$${\displaystyle a\cdot \frac{b^2}{a^2}=}$${\displaystyle \frac{b^2}{a}}$; ${\displaystyle {\left(\frac{a}{b}\right)}^{-1}\cdot b=}$${\displaystyle \frac{b}{a}\cdot b=}$${\displaystyle \frac{b^2}{a}}$;

${\displaystyle {\left(\frac{b^2}{a}\right)}^{-1}:{\left(\frac{b^2}{a}\right)}^2=}$${\displaystyle {\left(\frac{b^2}{a}\right)}^{-3}=}$${\displaystyle \frac{a^3}{b^6}=}$${\displaystyle a^3{b}^{-6}}$.