1382 Вправа 1382Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1382. Скороти дріб. а) (x² + 3x + 2)/(x² + 4x + 3); б) (z² + z − 6)/(z² − 2z − 15); в) (a² − 5a + 6)/(a² + 3a − 10).

а) ${\displaystyle \frac{x^2+3x+2}{x^2+4x+3}}$;

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

${\displaystyle x_1{x}_2=}$${\displaystyle 2}$ → ${\displaystyle x_1=}$${\displaystyle -1,x_2=}$${\displaystyle -2}$;

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

${\displaystyle x^2+4x+3}$: ${\displaystyle x_1+x_2=}$${\displaystyle -4}$;

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

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

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

б) ${\displaystyle \frac{z^2+z-6}{z^2-2z-15}}$;

${\displaystyle z^2+z-6}$: ${\displaystyle z_1+z_2=}$${\displaystyle -1}$;

${\displaystyle z_1{z}_2=}$${\displaystyle -6}$ → ${\displaystyle z_1=}$${\displaystyle 2,z_2=}$${\displaystyle -3}$;

${\displaystyle z^2+z-6=}$${\displaystyle \left(z-2\right)\left(z+3\right)}$;

${\displaystyle z^2-2z-15}$: ${\displaystyle z_1+z_2=}$${\displaystyle 2}$;

${\displaystyle z_1{z}_2=}$${\displaystyle -15}$ → ${\displaystyle z_1=}$${\displaystyle 5,z_2=}$${\displaystyle -3}$;

${\displaystyle z^2-2z-15=}$${\displaystyle \left(z-5\right)\left(z+3\right)}$;

${\displaystyle \frac{(z-2)\cancel{\left(z+3\right)}}{(z-5)\cancel{\left(z+3\right)}}=}$${\displaystyle \frac{z-2}{z-5}}$;

в) ${\displaystyle \frac{a^2-5a+6}{a^2+3a-10}}$;

${\displaystyle a^2-5a+6}$: ${\displaystyle a_1+a_2=}$${\displaystyle 5}$;

${\displaystyle a_1{a}_2=}$${\displaystyle 6}$ → ${\displaystyle a_1=}$${\displaystyle 3,a_2=}$${\displaystyle 2}$;

${\displaystyle a^2-5a+6=}$${\displaystyle \left(a-3\right)\left(a-2\right)}$;

${\displaystyle a^2+3a-10}$: ${\displaystyle a_1+a_2=}$${\displaystyle -3}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -10}$ → ${\displaystyle a_1=}$${\displaystyle 2,a_2=}$${\displaystyle -5}$;

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

${\displaystyle \frac{\cancel{\left(a-2\right)}(a-3)}{\cancel{\left(a-2\right)}(a+5)}=}$${\displaystyle \frac{a-3}{a+5}}$.