81 Вправа 81Алгебра 8 клас Бевз 2025 · § 2 · стор. 21

81. Скороти дріб.

а) ${\displaystyle \frac{7a^{2}b}{21a{b}^3}=}$${\displaystyle \frac{a\cdot \cancel{7ab}}{\cancel{7ab}\cdot 3b^2}=}$${\displaystyle \frac{a}{3b^2}}$;

б) ${\displaystyle \frac{35x{z}^5}{7x{z}^5}=}$${\displaystyle \frac{5\cdot \cancel{7x{z}^5}}{\cancel{7x{z}^5}}=}$${\displaystyle 5}$;

в) ${\displaystyle \frac{x^2-a^2}{3x^3-3a^{2}x}=}$${\displaystyle \frac{\cancel{\left(x^2-a^2\right)}}{3x\cancel{\left(x^2-a^2\right)}}=}$${\displaystyle \frac{1}{3x}}$;

г) ${\displaystyle \frac{5x^2-5xy}{5{(x-y)}^2}=}$${\displaystyle \frac{\cancel{5}x\cancel{\left(x-y\right)}}{\cancel{5}\cancel{\left(x-y\right)}(x-y)}=}$${\displaystyle \frac{x}{x-y}}$;

ґ) ${\displaystyle \frac{a^2-2ac+c^2}{{(a-c)}^3}=}$${\displaystyle \frac{\cancel{{\left(a-c\right)}^2}}{\cancel{{\left(a-c\right)}^2}(a-c)}=}$${\displaystyle \frac{1}{a-c}}$;

д) ${\displaystyle \frac{{(x+z)}^4}{x^2+2xz+z^2}=}$${\displaystyle \frac{{(x+z)}^2\cancel{{\left(x+z\right)}^2}}{\cancel{{\left(x+z\right)}^2}}=}$${\displaystyle {\left(x+z\right)}^2}$;

е) ${\displaystyle \frac{x^2-xz+z^2}{x^3+z^3}=}$${\displaystyle \frac{\cancel{\left(x^2-xz+z^2\right)}}{(x+z)\cancel{\left(x^2-xz+z^2\right)}}=}$${\displaystyle \frac{1}{x+z}}$.