155° Вправа 155Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 48

155. Спростіть вираз.

Знаменник-різницю квадратів розкладаємо на:

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

${\displaystyle \frac{2^{\mathrm{\setminus }x-3}}{3+x}-\frac{2x}{x^2-9}=}$${\displaystyle \frac{2\cdot \left(x-3\right)}{(x-3)(x+3)}-\frac{2x}{(x-3)(x+3)}=}$${\displaystyle \frac{2\cdot \left(x-3\right)-2x}{(x-3)(x+3)}=}$${\displaystyle \frac{2x-6-2x}{x^2-9}=}$${\displaystyle -\frac{6}{x^2-9}=}$${\displaystyle \frac{6}{9-x^2}}$;

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

${\displaystyle \frac{4x}{x^2-25}-\frac{4^{\mathrm{\setminus }x+5}}{x-5}=}$${\displaystyle \frac{4x}{(x-5)(x+5)}-\frac{4\cdot \left(x+5\right)}{(x-5)(x+5)}=}$${\displaystyle \frac{4x-4\cdot \left(x+5\right)}{(x-5)(x+5)}=}$${\displaystyle \frac{4x-4x-20}{x^2-25}=}$${\displaystyle -\frac{20}{x^2-25}}$;

3) ${\displaystyle 36-c^2=}$${\displaystyle (6-c)(6+c)}$;

${\displaystyle \frac{2^{\mathrm{\setminus }6-c}}{6+c}+\frac{2c}{36-c^2}=}$${\displaystyle \frac{2\cdot \left(6-c\right)}{(6-c)(6+c)}+\frac{2c}{(6-c)(6+c)}=}$${\displaystyle \frac{2\cdot \left(6-c\right)+2c}{(6-c)(6+c)}=}$${\displaystyle \frac{12-2c+2c}{36-c^2}=}$${\displaystyle \frac{12}{36-c^2}}$;

4) ${\displaystyle x^2-49=}$${\displaystyle (x-7)(x+7)}$;

${\displaystyle \frac{5^{\mathrm{\setminus }x-7}}{x+7}-\frac{10x}{x^2-49}=}$${\displaystyle \frac{5\cdot \left(x-7\right)}{(x-7)(x+7)}-\frac{10x}{(x-7)(x+7)}=}$${\displaystyle \frac{5\cdot \left(x-7\right)-10x}{(x-7)(x+7)}=}$${\displaystyle \frac{5x-35-10x}{(x-7)(x+7)}}$;${\displaystyle \frac{-5x-35}{(x-7)(x+7)}=}$${\displaystyle \frac{-5\cancel{\left(x+7\right)}}{(x-7)\cancel{\left(x+7\right)}}=}$${\displaystyle -\frac{5}{x-7}}$;

5) ${\displaystyle y^2-1=}$${\displaystyle (y-1)(y+1)}$;

${\displaystyle \frac{2^{\mathrm{\setminus }y-1}}{y+1}-\frac{1+2y}{y^2-1}=}$${\displaystyle \frac{2\cdot \left(y-1\right)}{y^2-1}-\frac{1+2y}{y^2-1}=}$${\displaystyle \frac{2\cdot \left(y-1\right)-(1+2y)}{y^2-1}=}$${\displaystyle \frac{2y-2-1-2y}{y^2-1}=}$${\displaystyle -\frac{3}{y^2-1}}$;

6) ${\displaystyle c^2-4=}$${\displaystyle (c-2)(c+2)}$;

${\displaystyle \frac{4}{c^2-4}-\frac{{4-c}^{\mathrm{\setminus }c+2}}{c-2}=}$${\displaystyle \frac{4}{c^2-4}-\frac{(4-c)\cdot \left(c+2\right)}{c^2-4}=}$${\displaystyle \frac{4-(4-c)\cdot \left(c+2\right)}{c^2-4}=}$${\displaystyle \frac{4-(4c+8-c^2-2c)}{c^2-4}}$;${\displaystyle \frac{4-2c-8+c^2}{c^2-4}=}$${\displaystyle \frac{c^2-2c-4}{c^2-4}}$.