102 Вправа 102Алгебра 8 клас Тарасенкова 2025 · § 2 · стор. 32

102. Скоротіть дріб.

Використовуємо різницю квадратів ${\displaystyle a^2-b^2=}$${\displaystyle (a-b)(a+b)}$.

1) ${\displaystyle \frac{x^4-x^2}{x^4+x^3}=}$${\displaystyle \frac{x^2(x^2-1)}{x^3(x+1)}=}$${\displaystyle \frac{\cancel{x^2}(x-1)\cancel{\left(x+1\right)}}{\cancel{x^2}\cdot x\cancel{\left(x+1\right)}}=}$${\displaystyle \frac{x-1}{x}}$;

2) ${\displaystyle \frac{y^3-16y}{4y^2+y^3}=}$${\displaystyle \frac{y(y^2-16)}{y^2(4+y)}=}$${\displaystyle \frac{\cancel{y}(y-4)\cancel{\left(y+4\right)}}{\cancel{y}\cdot y\cancel{\left(y+4\right)}}=}$${\displaystyle \frac{y-4}{y}}$;

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

4) ${\displaystyle \frac{4x^4-x^2}{6x+3}=}$${\displaystyle \frac{x^2(4x^2-1)}{3(2x+1)}=}$${\displaystyle \frac{x^2(2x-1)\cancel{\left(2x+1\right)}}{3\cancel{\left(2x+1\right)}}=}$${\displaystyle \frac{x^2\left(2x-1\right)}{3}}$;

5) ${\displaystyle \frac{a^3-25a{b}^2}{ab-5b^2}=}$${\displaystyle \frac{a(a^2-25b^2)}{b(a-5b)}=}$${\displaystyle \frac{a\cancel{\left(a-5b\right)}(a+5b)}{b\cancel{\left(a-5b\right)}}=}$${\displaystyle \frac{a\left(a+5b\right)}{b}}$;

6) ${\displaystyle \frac{4a^3-9a{c}^2}{2ac+3c^2}=}$${\displaystyle \frac{a(4a^2-9c^2)}{c(2a+3c)}=}$${\displaystyle \frac{a(2a-3c)\cancel{\left(2a+3c\right)}}{c\cancel{\left(2a+3c\right)}}=}$${\displaystyle \frac{a\left(2a-3c\right)}{c}}$.