1109 Вправа 1109Алгебра 8 клас Бевз 2025 · § 21 · стор. 207

1109. Скороти дріб з натуральними показниками степенів. а) 3ⁿ/3ⁿ⁺¹; б) 2³ⁿ/8ⁿ; в) aⁿ⁺²/aⁿ; г) xⁿ⁺³/x³; ґ) (a²ⁿ − 1)/(aⁿ + 1); д) (a²ⁿ − 2aⁿ + 1)/(aⁿ − 1).

а) ${\displaystyle \frac{3^n}{3^{n+1}}=}$${\displaystyle \frac{\cancel{3^n}}{\cancel{3^n}\cdot 3}=}$${\displaystyle \frac{1}{3}}$;

б) ${\displaystyle \frac{2^{3n}}{8^n}=}$${\displaystyle \frac{2^{3n}}{{\left(2^3\right)}^n}=}$${\displaystyle \frac{2^{3n}}{2^{3n}}=}$${\displaystyle 1}$;

в) ${\displaystyle \frac{a^{n+2}}{a^n}=}$${\displaystyle \frac{\cancel{a^n}\cdot a^2}{\cancel{a^n}}=}$${\displaystyle a^2}$;

г) ${\displaystyle \frac{x^{n+3}}{x^3}=}$${\displaystyle \frac{x^n\cdot \cancel{x^3}}{\cancel{x^3}}=}$${\displaystyle x^n}$;

ґ) ${\displaystyle a^{2n}-1=}$${\displaystyle {\left(a^n\right)}^2-1^2=}$${\displaystyle (a^n-1)(a^n+1)}$;

${\displaystyle \frac{a^{2n}-1}{a^n+1}=}$${\displaystyle \frac{(a^n-1)\cancel{\left(a^n+1\right)}}{\cancel{a^n+1}}=}$${\displaystyle a^n-1}$;

д) ${\displaystyle a^{2n}-2a^n+1=}$${\displaystyle {\left(a^n\right)}^2-2\cdot a^n\cdot 1+1^2=}$${\displaystyle {(a^n-1)}^2}$;

${\displaystyle \frac{a^{2n}-2a^n+1}{a^n-1}=}$${\displaystyle \frac{(a^n-1)\cancel{\left(a^n-1\right)}}{\cancel{a^n-1}}=}$${\displaystyle a^n-1}$.