1090 Вправа 1090Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1090. Обчисли значення дробу (x² + x − 6)/(x² + 4x + 3) при x = −1,1; x = 9; x = 11; x = 99.

${\displaystyle \frac{x^2+x-6}{x^2+4x+3}}$;

${\displaystyle x^2+x-6}$: ${\displaystyle x_1+x_2=}$${\displaystyle -1}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -6}$ → ${\displaystyle x_1=}$${\displaystyle 2,x_2=}$${\displaystyle -3}$;

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

${\displaystyle x=}$${\displaystyle -1,1}$: ${\displaystyle \frac{-3,1}{-0,1}=}$${\displaystyle 31}$;

${\displaystyle x=}$${\displaystyle 9}$: ${\displaystyle \frac{7}{10}=}$${\displaystyle 0,7}$; ${\displaystyle x=}$${\displaystyle 11}$: ${\displaystyle \frac{9}{12}=}$${\displaystyle 0,75}$; ${\displaystyle x=}$${\displaystyle 99}$: ${\displaystyle \frac{97}{100}=}$${\displaystyle 0,97}$.