823 Вправа 823Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 244

823. Знайдіть значення дробу:

1)

${\displaystyle x^2-2x-15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-15)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{2\pm \sqrt{64}}{2\cdot 1}=}$${\displaystyle \frac{2\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{2+8}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{2-8}{2}=}$${\displaystyle -3}$;

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

${\displaystyle \frac{x^2-2x-15}{x^2-9}=}$${\displaystyle \frac{\cancel{\left(x+3\right)}(x-5)}{(x-3)\cancel{\left(x+3\right)}}=}$${\displaystyle \frac{x-5}{x-3}}$;

якщо ${\displaystyle x=}$${\displaystyle -2}$: ${\displaystyle \frac{-2-5}{-2-3}=}$${\displaystyle \frac{-7}{-5}=}$${\displaystyle 1,4}$;

2)

${\displaystyle 3x^2-2x-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 3\cdot (-1)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{2\pm \sqrt{16}}{2\cdot 3}=}$${\displaystyle \frac{2\pm 4}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{2+4}{6}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{2-4}{6}=}$${\displaystyle \frac{-2}{6}=}$${\displaystyle -\frac{1}{3}}$;

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

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

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

якщо ${\displaystyle x=}$${\displaystyle 11}$: ${\displaystyle \frac{3\cdot 11+1}{11-1}=}$${\displaystyle \frac{34}{10}=}$${\displaystyle 3,4}$.