822 Вправа 822Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 244

822. Знайдіть значення дробу:

1)

${\displaystyle x^2-8x+12=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle 8}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 12}$;

${\displaystyle x_1=}$${\displaystyle 2}$; ${\displaystyle x_2=}$${\displaystyle 6}$;

${\displaystyle x^2-8x+12=}$${\displaystyle (x-2)(x-6)}$;

${\displaystyle \frac{x^2-36}{x^2-8x+12}=}$${\displaystyle \frac{\cancel{\left(x-6\right)}(x+6)}{(x-2)\cancel{\left(x-6\right)}}=}$${\displaystyle \frac{x+6}{x-2}}$;

якщо ${\displaystyle x=}$${\displaystyle 12}$: ${\displaystyle \frac{12+6}{12-2}=}$${\displaystyle \frac{18}{10}=}$${\displaystyle 1,8}$;

2)

${\displaystyle x^2-5x+6=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle 5}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 6}$;

${\displaystyle x_1=}$${\displaystyle 2}$; ${\displaystyle x_2=}$${\displaystyle 3}$;

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

${\displaystyle -2x^2+5x-2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 5^2-4\cdot (-2)\cdot (-2)=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm \sqrt{9}}{2\cdot (-2)}=}$${\displaystyle \frac{-5\pm 3}{-4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+3}{-4}=}$${\displaystyle \frac{-2}{-4}=}$${\displaystyle 0,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-3}{-4}=}$${\displaystyle 2}$;

${\displaystyle -2x^2+5x-2=}$${\displaystyle -2(x-2)(x-0,5)}$;

${\displaystyle \frac{x^2-5x+6}{-2x^2+5x-2}=}$${\displaystyle \frac{\cancel{\left(x-2\right)}(x-3)}{-2\cancel{\left(x-2\right)}(x-0,5)}=}$${\displaystyle \frac{x-3}{-2(x-0,5)}=}$${\displaystyle \frac{x-3}{1-2x}}$;

якщо ${\displaystyle x=}$${\displaystyle -0,5}$: ${\displaystyle \frac{-0,5-3}{1-2\cdot (-0,5)}=}$${\displaystyle \frac{-3,5}{1+1}=}$${\displaystyle \frac{-3,5}{2}=}$${\displaystyle -1,75}$.