1099 Вправа 1099Алгебра 8 клас Бевз 2025 · § 21 · стор. 206

1099. Спрости вираз. а) (2a² − 10a)/(a² − 2a − 3) · (a² + 2a − 15)/(a² − 25) − (a − 1)/(a + 1); б) (5/(x² + 7x + 12) − 1/(x² + 5x + 6) + x/(x² + 6x + 8)) · (x + 2)(x + 3)/((x + 1)(x + 6)).

а) розкладаємо на множники:

${\displaystyle 2a^2-10a=}$${\displaystyle 2a(a-5)}$;

${\displaystyle a^2-25=}$${\displaystyle (a-5)(a+5)}$;

${\displaystyle a^2-2a-3}$: ${\displaystyle a_1+a_2=}$${\displaystyle 2}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -3}$ → ${\displaystyle a_1=}$${\displaystyle 3,a_2=}$${\displaystyle -1}$;

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

${\displaystyle a^2+2a-15}$: ${\displaystyle a_1+a_2=}$${\displaystyle -2}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -15}$ → ${\displaystyle a_1=}$${\displaystyle 3,a_2=}$${\displaystyle -5}$;

${\displaystyle a^2+2a-15=}$${\displaystyle \left(a-3\right)\left(a+5\right)}$;

1) ${\displaystyle \frac{2a^2-10a}{a^2-2a-3}\cdot \frac{a^2+2a-15}{a^2-25}}$=

=${\displaystyle \frac{2a\cancel{\left(a-5\right)}}{\cancel{\left(a-3\right)}(a+1)}\cdot \frac{\cancel{\left(a+5\right)}\cancel{\left(a-3\right)}}{\cancel{\left(a-5\right)}\cancel{\left(a+5\right)}}=}$${\displaystyle \frac{2a}{a+1}}$;

2) ${\displaystyle \frac{2a}{a+1}-\frac{a-1}{a+1}=}$${\displaystyle \frac{a+1}{a+1}=}$${\displaystyle 1}$;

б) знаменники: ${\displaystyle (x+3)(x+4)}$, ${\displaystyle (x+2)(x+3)}$, ${\displaystyle (x+2)(x+4)}$;

1) ${\displaystyle \frac{5^{\mathrm{\setminus }x+2}}{(x+3)(x+4)}-\frac{1^{\mathrm{\setminus }x+4}}{(x+2)(x+3)}+\frac{x^{\mathrm{\setminus }x+3}}{(x+2)(x+4)}}$=

=${\displaystyle \frac{5x+10-x-4+x^2+3x}{(x+2)(x+3)(x+4)}=}$${\displaystyle \frac{\left(x+1\right)\left(x+6\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}}$;

${\displaystyle x^2+7x+6=}$${\displaystyle (x+1)(x+6)}$ (${\displaystyle x_1+x_2=}$${\displaystyle -7}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 6}$);

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