386° Вправа 386Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 108

386. Спростіть вираз:

${\displaystyle A^n\cdot A^m=}$${\displaystyle A^{n+m}}$; ${\displaystyle A^n:A^m=}$${\displaystyle A^{n-m}}$.

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

2) ${\displaystyle {(a-b)}^{-5}\cdot {(a-b)}^3=}$${\displaystyle {(a-b)}^{-5+3}=}$${\displaystyle {(a-b)}^{-2}=}$${\displaystyle \frac{1}{{\left(a-b\right)}^2}}$;

3) ${\displaystyle {(x-y)}^{-4}\cdot {(x-y)}^{-6}=}$${\displaystyle {(x-y)}^{-4+\left(-6\right)}=}$${\displaystyle \frac{1}{{\left(x-y\right)}^{10}}}$;

4) ${\displaystyle \frac{{(y+4)}^{-2}}{{(y+4)}^3}=}$${\displaystyle {(y+4)}^{-2-3}=}$${\displaystyle {(y+4)}^{-5}=}$${\displaystyle \frac{1}{{\left(y+4\right)}^5}}$;

5) ${\displaystyle \frac{{(c+1)}^{-2}}{{(c+1)}^{-4}}=}$${\displaystyle {(c+1)}^{-2-\left(-4\right)}=}$${\displaystyle {\left(c+1\right)}^2}$;

6) ${\displaystyle \frac{x^5\cdot {(x+y)}^{-3}}{{(x+y)}^{-5}\cdot x}=}$${\displaystyle x^{5-1}\cdot {(x+y)}^{-3-\left(-5\right)}=}$${\displaystyle x^4{\left(x+y\right)}^2}$;

7) ${\displaystyle \frac{m^6}{a^{-8}}:\frac{m^{-2}}{a^3}=}$${\displaystyle \frac{m^6\cdot a^3}{a^{-8}\cdot m^{-2}}=}$${\displaystyle m^{6+2}\cdot a^{3+8}=}$${\displaystyle m^8{a}^{11}}$;

8) ${\displaystyle \frac{4{(a+1)}^{-18}}{c^9}:\frac{c^{-2}}{{(a+1)}^{-22}}=}$${\displaystyle \frac{4{(a+1)}^{-18}\cdot {(a+1)}^{-22}}{c^9\cdot c^{-2}}=}$${\displaystyle \frac{4{(a+1)}^{-18-22}}{c^{9-2}}=}$${\displaystyle \frac{4}{c^7{\left(a+1\right)}^{40}}}$;

9) ${\displaystyle \frac{{(x-y)}^{-2}\cdot y^{-5}\cdot {(x-y)}^5}{y\cdot {(x-y)}^7\cdot y^{-3}}=}$${\displaystyle {(x-y)}^{-2+5-7}\cdot y^{-5-1+3}=}$${\displaystyle {(x-y)}^{-4}{y}^{-3}=}$${\displaystyle \frac{1}{y^3{\left(x-y\right)}^4}}$.