43 Вправа 43 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

43. Скоротіть дріб:

1) Розкладемо чисельник на множники:

${\displaystyle 3x^2+5x+2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 5^2-4\cdot 3\cdot 2=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm \sqrt{1}}{2\cdot 3}=}$${\displaystyle \frac{-5\pm 1}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+1}{6}=}$${\displaystyle \frac{-4}{6}=}$${\displaystyle -\frac{2}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-1}{6}=}$${\displaystyle -1}$;

${\displaystyle 3x^2+5x+2=}$${\displaystyle 3(x+1)(x+\frac{2}{3})=}$${\displaystyle (x+1)(3x+2)}$;

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

2) ${\displaystyle x^2-11x-26=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle 11}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -26}$;

${\displaystyle x_1=}$${\displaystyle -2}$; ${\displaystyle x_2=}$${\displaystyle 13}$;

${\displaystyle x^2-11x-26=}$${\displaystyle (x+2)(x-13)}$; ${\displaystyle x^2-4=}$${\displaystyle (x-2)(x+2)}$;

${\displaystyle \frac{x^2-11x-26}{x^2-4}=}$${\displaystyle \frac{\cancel{\left(x+2\right)}(x-13)}{(x-2)\cancel{\left(x+2\right)}}=}$${\displaystyle \frac{x-13}{x-2}}$;

3) ${\displaystyle 6x^2-17x+5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-17)}^2-4\cdot 6\cdot 5=}$${\displaystyle 289-120=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{17\pm \sqrt{169}}{2\cdot 6}=}$${\displaystyle \frac{17\pm 13}{12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{17+13}{12}=}$${\displaystyle \frac{30}{12}=}$${\displaystyle 2,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{17-13}{12}=}$${\displaystyle \frac{4}{12}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle 6x^2-17x+5=}$${\displaystyle 6(x-2,5)(x-\frac{1}{3})=}$${\displaystyle 2(x-2,5)\cdot 3(x-\frac{1}{3})=}$${\displaystyle (2x-5)(3x-1)}$;

${\displaystyle \frac{6x^2-17x+5}{3x-1}=}$${\displaystyle \frac{(2x-5)\cancel{\left(3x-1\right)}}{\cancel{\left(3x-1\right)}}=}$${\displaystyle 2x-5}$;

Виправлено: у розв’язнику записано 6x² – 17x + 5 = –6(x – 2,5)(x – 1/3) і відповідь 5 – 2x — це помилка в знаку; правильно 2x – 5.

4) ${\displaystyle x^2-6x+8=}$${\displaystyle 0}$: ${\displaystyle x_1=}$${\displaystyle 2}$; ${\displaystyle x_2=}$${\displaystyle 4}$; ${\displaystyle x^2-6x+8=}$${\displaystyle (x-2)(x-4)}$;

${\displaystyle x^2-3x-4=}$${\displaystyle 0}$: ${\displaystyle x_1=}$${\displaystyle -1}$; ${\displaystyle x_2=}$${\displaystyle 4}$; ${\displaystyle x^2-3x-4=}$${\displaystyle (x+1)(x-4)}$;

${\displaystyle \frac{x^2-6x+8}{x^2-3x-4}=}$${\displaystyle \frac{(x-2)\cancel{\left(x-4\right)}}{(x+1)\cancel{\left(x-4\right)}}=}$${\displaystyle \frac{x-2}{x+1}}$.