1094 Вправа 1094Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1094. Знайди суму і різницю дробів. а) 1/(x² − 2x − 3) і 3/(x² − x − 6); б) 1/(2x² + 5x − 3) і 1/(2x² − 7x + 3).

а) ${\displaystyle \frac{1}{x^2-2x-3}}$ і ${\displaystyle \frac{3}{x^2-x-6}}$;

${\displaystyle x^2-2x-3}$: ${\displaystyle x_1+x_2=}$${\displaystyle 2}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -3}$ → ${\displaystyle x_1=}$${\displaystyle 3,x_2=}$${\displaystyle -1}$;

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

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

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

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

спільний знаменник: ${\displaystyle (x-3)(x+1)(x+2)}$;

сума: ${\displaystyle \frac{1^{\mathrm{\setminus }x+2}}{(x-3)(x+1)}+\frac{3^{\mathrm{\setminus }x+1}}{(x-3)(x+2)}}$=

=${\displaystyle \frac{x+2+3x+3}{(x-3)(x+1)(x+2)}=}$${\displaystyle \frac{4x+5}{\left(x-3\right)\left(x+1\right)\left(x+2\right)}}$;

різниця: ${\displaystyle \frac{x+2-3x-3}{(x-3)(x+1)(x+2)}}$=

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

б) ${\displaystyle \frac{1}{2x^2+5x-3}}$ і ${\displaystyle \frac{1}{2x^2-7x+3}}$;

${\displaystyle 2x^2+5x-3}$: ${\displaystyle D=}$${\displaystyle 5^2-4\cdot 2\cdot (-3)=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm 7}{4}}$;

${\displaystyle x_1=}$${\displaystyle 0,5}$;

${\displaystyle x_2=}$${\displaystyle -3}$;

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

${\displaystyle 2x^2-7x+3}$: ${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 2\cdot 3=}$${\displaystyle 25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{7\pm 5}{4}}$;

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

${\displaystyle x_2=}$${\displaystyle 0,5}$;

${\displaystyle 2x^2-7x+3=}$${\displaystyle 2(x-3)(x-0,5)=}$${\displaystyle \left(x-3\right)\left(2x-1\right)}$;

спільний знаменник: ${\displaystyle (2x-1)(x+3)(x-3)}$;

сума: ${\displaystyle \frac{1^{\mathrm{\setminus }x-3}}{(2x-1)(x+3)}+\frac{1^{\mathrm{\setminus }x+3}}{(2x-1)(x-3)}}$=

=${\displaystyle \frac{x-3+x+3}{(2x-1)(x+3)(x-3)}=}$${\displaystyle \frac{2x}{\left(2x-1\right)\left(x^2-9\right)}}$;

різниця: ${\displaystyle \frac{x-3-x-3}{(2x-1)(x+3)(x-3)}=}$${\displaystyle -\frac{6}{\left(2x-1\right)\left(x^2-9\right)}}$.