44 Вправа 44 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

44. Знайдіть значення дробу:

1) ${\displaystyle \frac{7x^2-6x-1}{7x+1}}$, якщо ${\displaystyle x=}$${\displaystyle -28}$;

${\displaystyle 7x^2-6x-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-6)}^2-4\cdot 7\cdot (-1)=}$${\displaystyle 36+28=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{6\pm \sqrt{64}}{2\cdot 7}=}$${\displaystyle \frac{6\pm 8}{14}}$;

${\displaystyle x_1=}$${\displaystyle \frac{6+8}{14}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{6-8}{14}=}$${\displaystyle \frac{-2}{14}=}$${\displaystyle -\frac{1}{7}}$;

${\displaystyle 7x^2-6x-1=}$${\displaystyle 7(x+\frac{1}{7})(x-1)=}$${\displaystyle (7x+1)(x-1)}$;

${\displaystyle \frac{7x^2-6x-1}{7x+1}=}$${\displaystyle \frac{\cancel{\left(7x+1\right)}(x-1)}{\cancel{\left(7x+1\right)}}=}$${\displaystyle x-1}$;

якщо ${\displaystyle x=}$${\displaystyle -28}$, то ${\displaystyle -28-1=}$${\displaystyle -29}$;

2) ${\displaystyle \frac{6x^2-11x-2}{1+6x}}$, якщо ${\displaystyle x=}$${\displaystyle -0,5}$;

${\displaystyle 6x^2-11x-2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-11)}^2-4\cdot 6\cdot (-2)=}$${\displaystyle 121+48=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{11\pm \sqrt{169}}{2\cdot 6}=}$${\displaystyle \frac{11\pm 13}{12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{11+13}{12}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-13}{12}=}$${\displaystyle \frac{-2}{12}=}$${\displaystyle -\frac{1}{6}}$;

${\displaystyle 6x^2-11x-2=}$${\displaystyle 6(x+\frac{1}{6})(x-2)=}$${\displaystyle (6x+1)(x-2)}$;

${\displaystyle \frac{6x^2-11x-2}{1+6x}=}$${\displaystyle \frac{\cancel{\left(6x+1\right)}(x-2)}{\cancel{\left(1+6x\right)}}=}$${\displaystyle x-2}$;

якщо ${\displaystyle x=}$${\displaystyle -0,5}$, то ${\displaystyle -0,5-2=}$${\displaystyle -2,5}$.