401 Вправа 401Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 111

401. Спростіть вираз:

1) ${\displaystyle \frac{16x^{-5}}{y^{-12}}\cdot {\left(\frac{y^{16}}{2x^{-4}{z}^9}\right)}^5=}$${\displaystyle \frac{16x^{-5}}{y^{-12}}\cdot \frac{y^{80}}{32x^{-20}{z}^{45}}=}$${\displaystyle \frac{16}{32}{x}^{-5+20}{y}^{12+80}{z}^{-45}=}$${\displaystyle 0,5x^{15}{y}^{92}{z}^{-45}}$;

2) ${\displaystyle {\left(\frac{x^{-1}}{y^{-4}{z}^7}\right)}^{-5}:{\left(\frac{x}{y^{10}{z}^{-20}}\right)}^3=}$${\displaystyle {\left(x^{-1}{y}^4{z}^{-7}\right)}^{-5}:{\left(x{y}^{-10}{z}^{20}\right)}^3=}$${\displaystyle x^5{y}^{-20}{z}^{35}:x^3{y}^{-30}{z}^{60}=}$${\displaystyle x^{5-3}{y}^{-20+30}{z}^{35-60}=}$${\displaystyle x^2{y}^{10}{z}^{-25}}$;

3) ${\displaystyle {\left(\frac{2x^3}{y^4{z}^{-5}}\right)}^2\cdot {\left(\frac{x^8}{0,25y^{-9}}\right)}^{-2}=}$${\displaystyle 4x^6{y}^{-8}{z}^{10}\cdot {\left(4x^8{y}^9\right)}^{-2}=}$${\displaystyle 4x^6{y}^{-8}{z}^{10}\cdot \frac{1}{16}{x}^{-16}{y}^{-18}=}$${\displaystyle 0,25x^{-10}{y}^{-26}{z}^{10}}$;

4) ${\displaystyle {\left(\frac{2x^{-2}}{5y^6{z}^{11}}\right)}^{-2}:{\left(\frac{0,2x^{-2}}{y^{-5}{z}^0}\right)}^{-3}=}$${\displaystyle {(0,4x^{-2}{y}^{-6}{z}^{-11})}^{-2}:{(0,2x^{-2}{y}^5)}^{-3}=}$${\displaystyle 6,25x^4{y}^{12}{z}^{22}:125x^6{y}^{-15}=}$${\displaystyle 0,05x^{-2}{y}^{27}{z}^{22}}$;

5) ${\displaystyle {\left(\frac{x^{-15}}{6y^6}\right)}^{-2}:{\left(\frac{x^{-20}}{3y^9}\right)}^{-2}=}$${\displaystyle {(\frac{x^{-15}}{6y^6}\cdot \frac{3y^9}{x^{-20}})}^{-2}=}$${\displaystyle {\left(\frac{x^5{y}^3}{2}\right)}^{-2}=}$${\displaystyle 4x^{-10}{y}^{-6}}$;

6) ${\displaystyle {\left(\frac{1\frac{2}{3}{a}^{-2}}{b^6}\right)}^{-1}:{\left(\frac{0,6a^{-8}}{b^{-3}}\right)}^2=}$${\displaystyle {\left(\frac{5}{3}{a}^{-2}{b}^{-6}\right)}^{-1}:{(0,6a^{-8}{b}^3)}^2=}$${\displaystyle 0,6a^2{b}^6:0,36a^{-16}{b}^6=}$${\displaystyle \frac{0,6}{0,36}{a}^{2+16}=}$${\displaystyle \frac{5}{3}{a}^{18}}$.