5 Вправа 5 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 316

5. Зведіть до найменшого спільного знаменника дроби:

1) ${\displaystyle \frac{1}{x-3}}$, ${\displaystyle \frac{1}{3x-9}}$, ${\displaystyle \frac{1}{18-6x}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }6}}{x-3}}$, ${\displaystyle \frac{1^{\mathrm{\setminus }2}}{3(x-3)}}$, ${\displaystyle \frac{1}{6(3-x)}=}$${\displaystyle \frac{1^{\mathrm{\setminus }-1}}{-6(x-3)}}$;

${\displaystyle \frac{6}{6\left(x-3\right)}}$, ${\displaystyle \frac{2}{6\left(x-3\right)}}$, ${\displaystyle \frac{-1}{6\left(x-3\right)}}$;

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

${\displaystyle \frac{1^{\mathrm{\setminus }x+1}}{x}}$, ${\displaystyle \frac{1}{x(x+1)}}$, ${\displaystyle \frac{1^{\mathrm{\setminus }x}}{x+1}}$;

${\displaystyle \frac{x+1}{x\left(x+1\right)}}$, ${\displaystyle \frac{1}{x\left(x+1\right)}}$, ${\displaystyle \frac{x}{x\left(x+1\right)}}$;

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

${\displaystyle \frac{1^{\mathrm{\setminus }x-2}}{x+2}}$, ${\displaystyle \frac{1}{(x-2)(x+2)}}$, ${\displaystyle \frac{1^{\mathrm{\setminus }x+2}}{x-2}}$;

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

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

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

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

${\displaystyle \frac{{\left(x-1\right)}^2}{\left(x-1\right)\left(x^3-1\right)}}$, ${\displaystyle \frac{x-1}{\left(x-1\right)\left(x^3-1\right)}}$, ${\displaystyle \frac{x^2+x+1}{\left(x-1\right)\left(x^3-1\right)}}$.