129 Вправа 129Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 40

129. Зведіть до спільного знаменника дроби.

1) ${\displaystyle \frac{1}{a^2-9}}$, ${\displaystyle \frac{1}{2a+6}}$, ${\displaystyle \frac{1}{3-a}}$;

${\displaystyle a^2-9=}$${\displaystyle (a-3)(a+3)}$; ${\displaystyle 2a+6=}$${\displaystyle 2(a+3)}$; ${\displaystyle 3-a=}$${\displaystyle -(a-3)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2}}{(a-3)(a+3)}=}$${\displaystyle \frac{1\cdot 2}{(a-3)(a+3)\cdot 2}=}$${\displaystyle \frac{2}{2\left(a^2-9\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }a-3}}{2(a+3)}=}$${\displaystyle \frac{1\cdot \left(a-3\right)}{2(a+3)\cdot \left(a-3\right)}=}$${\displaystyle \frac{a-3}{2\left(a^2-9\right)}}$;

${\displaystyle \frac{1}{3-a}=}$${\displaystyle \frac{{-1}^{\mathrm{\setminus }2\left(a+3\right)}}{a-3}=}$${\displaystyle \frac{-1\cdot 2\left(a+3\right)}{(a-3)\cdot 2\left(a+3\right)}=}$${\displaystyle \frac{-2\left(a+3\right)}{2\left(a^2-9\right)}}$;

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

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

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

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

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