1095 Вправа 1095Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1095. Знайди суму і різницю дробів. а) 2/(x² + 3x − 10) і 1/(x² − 7x + 10); б) 1/(6a² − 13a + 6) і 1/(3a² − 11a + 6).

а) ${\displaystyle \frac{2}{x^2+3x-10}}$ і ${\displaystyle \frac{1}{x^2-7x+10}}$;

${\displaystyle x^2+3x-10}$: ${\displaystyle x_1+x_2=}$${\displaystyle -3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -10}$ → ${\displaystyle x_1=}$${\displaystyle 2,x_2=}$${\displaystyle -5}$;

${\displaystyle x^2+3x-10=}$${\displaystyle \left(x-2\right)\left(x+5\right)}$;

${\displaystyle x^2-7x+10}$: ${\displaystyle x_1+x_2=}$${\displaystyle 7}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 10}$ → ${\displaystyle x_1=}$${\displaystyle 5,x_2=}$${\displaystyle 2}$;

${\displaystyle x^2-7x+10=}$${\displaystyle \left(x-5\right)\left(x-2\right)}$;

спільний знаменник: ${\displaystyle (x-2)(x+5)(x-5)}$;

сума: ${\displaystyle \frac{2^{\mathrm{\setminus }x-5}}{(x-2)(x+5)}+\frac{1^{\mathrm{\setminus }x+5}}{(x-5)(x-2)}}$=

=${\displaystyle \frac{2x-10+x+5}{(x-2)(x+5)(x-5)}=}$${\displaystyle \frac{3x-5}{\left(x-2\right)\left(x^2-25\right)}}$;

різниця: ${\displaystyle \frac{2x-10-x-5}{(x-2)(x+5)(x-5)}=}$${\displaystyle \frac{x-15}{\left(x-2\right)\left(x^2-25\right)}}$;

б) ${\displaystyle \frac{1}{6a^2-13a+6}}$ і ${\displaystyle \frac{1}{3a^2-11a+6}}$;

${\displaystyle 6a^2-13a+6}$: ${\displaystyle D=}$${\displaystyle {(-13)}^2-4\cdot 6\cdot 6=}$${\displaystyle 25}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{13\pm 5}{12}}$;

${\displaystyle a_1=}$${\displaystyle 1,5}$;

${\displaystyle a_2=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle 6a^2-13a+6=}$${\displaystyle 6(a-1,5)(a-\frac{2}{3})=}$${\displaystyle \left(2a-3\right)\left(3a-2\right)}$;

${\displaystyle 3a^2-11a+6}$: ${\displaystyle D=}$${\displaystyle {(-11)}^2-4\cdot 3\cdot 6=}$${\displaystyle 49}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{11\pm 7}{6}}$;

${\displaystyle a_1=}$${\displaystyle 3}$;

${\displaystyle a_2=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle 3a^2-11a+6=}$${\displaystyle 3(a-3)(a-\frac{2}{3})=}$${\displaystyle \left(a-3\right)\left(3a-2\right)}$;

спільний знаменник: ${\displaystyle (3a-2)(2a-3)(a-3)}$;

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

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

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

=${\displaystyle -\frac{a}{\left(3a-2\right)\left(2a-3\right)\left(a-3\right)}}$.