132* Вправа 132Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 40

132. Зведіть до спільного знаменника дроби.

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

Розкладаємо знаменники (групування):

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

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

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

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

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

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

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

${\displaystyle \frac{1}{x^4-1}}$ — без змін.

2) ${\displaystyle \frac{1}{x^2+3x+2}}$, ${\displaystyle \frac{1}{x^2+5x+6}}$, ${\displaystyle \frac{1}{x^2+7x+10}}$;

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

${\displaystyle x^2+5x+6=}$${\displaystyle x^2+2x+3x+6=}$${\displaystyle x(x+2)+3(x+2)=}$${\displaystyle (x+2)(x+3)}$;

${\displaystyle x^2+7x+10=}$${\displaystyle x^2+2x+5x+10=}$${\displaystyle x(x+2)+5(x+2)=}$${\displaystyle (x+2)(x+5)}$;

${\displaystyle \frac{1}{(x+1)(x+2)}=}$${\displaystyle \frac{x^2+8x+15}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+5\right)}}$;

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

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

(Перевірка чисельників: ${\displaystyle (x+3)(x+5)=}$${\displaystyle x^2+8x+15}$; ${\displaystyle (x+1)(x+5)=}$${\displaystyle x^2+6x+5}$; ${\displaystyle (x+1)(x+3)=}$${\displaystyle x^2+4x+3}$.)