131 Вправа 131Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 40

131. Скоротіть дроби, а потім зведіть їх до спільного знаменника:

1) Скорочуємо:

${\displaystyle \frac{ax-2a}{2b-bx}=}$${\displaystyle \frac{a(x-2)}{b(2-x)}=}$${\displaystyle \frac{a\cancel{\left(x-2\right)}}{-b\cancel{\left(x-2\right)}}=}$${\displaystyle -\frac{a}{b}}$;

${\displaystyle \frac{4a^2+8ab+4b^2}{{(a+b)}^3}=}$${\displaystyle \frac{4\cancel{{\left(a+b\right)}^2}}{\cancel{{\left(a+b\right)}^2}(a+b)}=}$${\displaystyle \frac{4}{a+b}}$;

${\displaystyle -\frac{a\cdot \left(a+b\right)}{b\cdot \left(a+b\right)}=}$${\displaystyle \frac{-a\left(a+b\right)}{b\left(a+b\right)}}$;

${\displaystyle \frac{4\cdot b}{(a+b)\cdot b}=}$${\displaystyle \frac{4b}{b\left(a+b\right)}}$;

2) Скорочуємо:

${\displaystyle \frac{x^2+10x+25}{3x+15}=}$${\displaystyle \frac{(x+5)\cancel{\left(x+5\right)}}{3\cancel{\left(x+5\right)}}=}$${\displaystyle \frac{x+5}{3}}$;

${\displaystyle \frac{x^2-6x+9}{x^2-9}=}$${\displaystyle \frac{\cancel{\left(x-3\right)}(x-3)}{\cancel{\left(x-3\right)}(x+3)}=}$${\displaystyle \frac{x-3}{x+3}}$;

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

${\displaystyle \frac{(x-3)\cdot 3}{(x+3)\cdot 3}=}$${\displaystyle \frac{3x-9}{3\left(x+3\right)}}$.