128 Вправа 128Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 40

128. Зведіть до спільного знаменника дроби.

Різниця і сума кубів: ${\displaystyle a^3-b^3=}$${\displaystyle (a-b)(a^2+ab+b^2)}$; ${\displaystyle a^3+b^3=}$${\displaystyle (a+b)(a^2-ab+b^2)}$.

1) ${\displaystyle \frac{1}{x-1}}$, ${\displaystyle \frac{1}{x^3-1}}$, ${\displaystyle \frac{1}{x^2+x+1}}$;

${\displaystyle x^3-1=}$${\displaystyle (x-1)(x^2+x+1)}$;

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

${\displaystyle \frac{1}{x^3-1}}$ — без змін;

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

2) ${\displaystyle \frac{1}{x+2}}$, ${\displaystyle \frac{1}{x^3+8}}$, ${\displaystyle \frac{1}{x^2-4}}$;

${\displaystyle x^3+8=}$${\displaystyle (x+2)(x^2-2x+4)}$; ${\displaystyle x^2-4=}$${\displaystyle (x-2)(x+2)}$;

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

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

${\displaystyle \frac{1}{(x-2)(x+2)}=}$${\displaystyle \frac{x^2-2x+4}{\left(x^2-4\right)\left(x^2-2x+4\right)}}$.