387° Вправа 387Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 108

387. Спростіть вираз:

1) ${\displaystyle {(m+n)}^3\cdot {(m+n)}^{-3}=}$${\displaystyle {(m+n)}^{3+\left(-3\right)}=}$${\displaystyle {(m+n)}^0=}$${\displaystyle 1}$;

2) ${\displaystyle {(5+a)}^{-8}\cdot {(5+a)}^5=}$${\displaystyle {(5+a)}^{-8+5}=}$${\displaystyle {(5+a)}^{-3}=}$${\displaystyle \frac{1}{{\left(5+a\right)}^3}}$;

3) ${\displaystyle \frac{{(p+m)}^{-1}}{{(p+m)}^2}=}$${\displaystyle {(p+m)}^{-1-2}=}$${\displaystyle {(p+m)}^{-3}=}$${\displaystyle \frac{1}{{\left(p+m\right)}^3}}$;

4) ${\displaystyle {\left(\frac{m}{n}\right)}^{-1}\cdot {\left(\frac{m}{n}\right)}^{-5}=}$${\displaystyle {\left(\frac{m}{n}\right)}^{-1+\left(-5\right)}=}$${\displaystyle {\left(\frac{m}{n}\right)}^{-6}=}$${\displaystyle {\left(\frac{n}{m}\right)}^6=}$${\displaystyle \frac{n^6}{m^6}}$;

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

6) ${\displaystyle \frac{x^{-3}}{{(y+1)}^0}\cdot \frac{{(x+2)}^0}{x^{-1}}=}$${\displaystyle \frac{x^{-3}}{1}\cdot \frac{1}{x^{-1}}=}$${\displaystyle x^{-3-\left(-1\right)}=}$${\displaystyle x^{-2}=}$${\displaystyle \frac{1}{x^2}}$;

7) ${\displaystyle \frac{{(y+5)}^{-4}}{y^{10}}\cdot \frac{y^{-15}}{{(y+5)}^{-8}}=}$${\displaystyle {(y+5)}^{-4-\left(-8\right)}\cdot y^{-15-10}=}$${\displaystyle {(y+5)}^4\cdot y^{-25}=}$${\displaystyle \frac{{\left(y+5\right)}^4}{y^{25}}}$;

8) ${\displaystyle \frac{{(n+4m)}^{-1}}{p^4}\cdot \frac{p^3}{{(n+4m)}^0}:\frac{{(n+4m)}^{-10}}{p}=}$${\displaystyle \frac{{(n+4m)}^{-1}\cdot p^3\cdot p}{p^4\cdot 1\cdot {(n+4m)}^{-10}}=}$${\displaystyle {(n+4m)}^{-1+10}\cdot p^{3+1-4}=}$${\displaystyle {\left(n+4m\right)}^9}$.