405 Вправа 405Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 112

405. Винесіть за дужки y⁻²:

Кожен доданок ділимо на ${\displaystyle y^{-2}}$: ${\displaystyle y^n:y^{-2}=}$${\displaystyle y^{n+2}}$.

1) ${\displaystyle y^{-8}+y^{-4}=}$${\displaystyle y^{-2}\left(y^{-6}+y^{-2}\right)}$;

2) ${\displaystyle y^{-10}-y^2=}$${\displaystyle y^{-2}\left(y^{-8}-y^4\right)}$;

3) ${\displaystyle y^4+8y^{-4}=}$${\displaystyle y^{-2}\left(y^6+8y^{-2}\right)}$;

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

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

6) ${\displaystyle x{y}^{-9}+9y^{-2}=}$${\displaystyle y^{-2}\left(x{y}^{-7}+9\right)}$;

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

8) ${\displaystyle y^2{x}^{-2}-5=}$${\displaystyle y^{-2}\left(x^{-2}{y}^4-5y^2\right)}$;

9) ${\displaystyle x{y}^{-1}-\frac{6}{y^3}+y^6+3=}$${\displaystyle y^{-2}\left(xy-6y^{-1}+y^8+3y^2\right)}$.