1081 Вправа 1081Алгебра 8 клас Бевз 2025 · § 21 · стор. 204

1081. Скороти дріб. а) (x + 5)/(x² + 7x + 10); б) (x − 3)/(x² − 2x − 3); в) (12 + 3x)/(x² + 5x + 4).

а) ${\displaystyle \frac{x+5}{x^2+7x+10}}$;

${\displaystyle x^2+7x+10}$: ${\displaystyle x_1+x_2=}$${\displaystyle -7}$;

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

${\displaystyle x^2+7x+10=}$${\displaystyle \left(x+2\right)\left(x+5\right)}$;

${\displaystyle \frac{\cancel{x+5}}{(x+2)\cancel{\left(x+5\right)}}=}$${\displaystyle \frac{1}{x+2}}$;

(${\displaystyle x\ne -5}$);

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

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

(${\displaystyle x\ne 3}$);

в) ${\displaystyle \frac{12+3x}{x^2+5x+4}}$;

${\displaystyle x^2+5x+4}$: ${\displaystyle x_1+x_2=}$${\displaystyle -5}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 4}$ → ${\displaystyle x_1=}$${\displaystyle -1,x_2=}$${\displaystyle -4}$;

${\displaystyle x^2+5x+4=}$${\displaystyle \left(x+1\right)\left(x+4\right)}$;

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

(${\displaystyle x\ne -4}$).