1080 Вправа 1080Алгебра 8 клас Бевз 2025 · § 21 · стор. 204

1080. Скороти дріб. а) (x² − 2x − 3)/(x + 1); б) (x − 4)/(x² − x − 12); в) (2x − 10)/(x² − 3x − 10).

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

${\displaystyle x^2-2x-3}$: ${\displaystyle x_1+x_2=}$${\displaystyle 2}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -3}$ → ${\displaystyle x_1=}$${\displaystyle 3,x_2=}$${\displaystyle -1}$;

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

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

(${\displaystyle x\ne -1}$);

б) ${\displaystyle \frac{x-4}{x^2-x-12}}$;

${\displaystyle x^2-x-12}$: ${\displaystyle x_1+x_2=}$${\displaystyle 1}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -12}$ → ${\displaystyle x_1=}$${\displaystyle 4,x_2=}$${\displaystyle -3}$;

${\displaystyle x^2-x-12=}$${\displaystyle \left(x-4\right)\left(x+3\right)}$;

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

(${\displaystyle x\ne 4}$);

в) ${\displaystyle \frac{2x-10}{x^2-3x-10}}$;

${\displaystyle x^2-3x-10}$: ${\displaystyle x_1+x_2=}$${\displaystyle 3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -10}$ → ${\displaystyle x_1=}$${\displaystyle 5,x_2=}$${\displaystyle -2}$;

${\displaystyle x^2-3x-10=}$${\displaystyle \left(x-5\right)\left(x+2\right)}$;

${\displaystyle \frac{2\cancel{\left(x-5\right)}}{\cancel{\left(x-5\right)}(x+2)}=}$${\displaystyle \frac{2}{x+2}}$;

(${\displaystyle x\ne 5}$).