400 Вправа 400Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 110

400. Спростіть вираз:

1) ${\displaystyle \frac{1}{9n^{45}{m}^{-35}}\cdot \frac{m^{-20}}{n^{-30}}=}$${\displaystyle \frac{m^{-20}}{9n^{45}{m}^{-35}{n}^{-30}}=}$${\displaystyle \frac{m^{-20+35}}{9n^{45-30}}=}$${\displaystyle \frac{m^{15}}{9n^{15}}}$;

2) ${\displaystyle 5b{a}^0\cdot \frac{a^5}{4b^{-3}}=}$${\displaystyle \frac{5b\cdot 1\cdot a^5}{4b^{-3}}=}$${\displaystyle \frac{5}{4}{a}^5{b}^{1+3}=}$${\displaystyle 1,25a^5{b}^4}$;

3) ${\displaystyle \frac{2b^{-8}}{3a^2}\cdot 0,6b^{-1}{a}^{-9}=}$${\displaystyle \frac{2\cdot 0,6}{3}\cdot a^{-2-9}{b}^{-8-1}=}$${\displaystyle 0,4a^{-11}{b}^{-9}}$;

4) ${\displaystyle 5xyz\cdot \frac{2x^{-2}}{5y^6{z}^{-11}}=}$${\displaystyle \frac{\cancel{5}\cdot 2x^{1-2}{y}^{1-6}{z}^{1+11}}{\cancel{5}}=}$${\displaystyle 2x^{-1}{y}^{-5}{z}^{12}=}$${\displaystyle \frac{2z^{12}}{x{y}^5}}$;

5) ${\displaystyle \frac{2x^{-4}}{y^{-2}}\cdot 0,1x{y}^{-5}=}$${\displaystyle 2\cdot 0,1\cdot x^{-4+1}{y}^{2-5}=}$${\displaystyle 0,2x^{-3}{y}^{-3}}$;

6) ${\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}}$;

7) ${\displaystyle {({(x^0\cdot x^{-7})}^{-5}\cdot x^{-1})}^{10}=}$${\displaystyle {({\left(x^{-7}\right)}^{-5}\cdot x^{-1})}^{10}=}$${\displaystyle {(x^{35}\cdot x^{-1})}^{10}=}$${\displaystyle {\left(x^{34}\right)}^{10}=}$${\displaystyle x^{340}}$;

8) ${\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}$;

9) ${\displaystyle ({(x^{-9}:x^{-1})}^{-1}:x^7):x^0=}$${\displaystyle ({\left(x^{-8}\right)}^{-1}:x^7):1=}$${\displaystyle x^8:x^7=}$${\displaystyle x}$;

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