159 Вправа 159Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 49

159. Виконайте дії.

Після віднімання (додавання) чисельник розкладаємо за формулами суми й різниці кубів: ${\displaystyle a^3\pm b^3=}$${\displaystyle (a\pm b)(a^2\mp ab+b^2)}$ — і скорочуємо.

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

2) ${\displaystyle \frac{q^3}{p-q}-\frac{p^3}{p-q}=}$${\displaystyle \frac{q^3-p^3}{p-q}=}$${\displaystyle \frac{-\cancel{\left(p-q\right)}(p^2+pq+q^2)}{\cancel{p-q}}=}$${\displaystyle -p^2-pq-q^2}$;

3) ${\displaystyle \frac{2x^3}{x+y}-\frac{x^3-y^3}{x+y}=}$${\displaystyle \frac{2x^3-x^3+y^3}{x+y}=}$${\displaystyle \frac{x^3+y^3}{x+y}=}$${\displaystyle \frac{\cancel{\left(x+y\right)}(x^2-xy+y^2)}{\cancel{x+y}}=}$${\displaystyle x^2-xy+y^2}$;

4) ${\displaystyle \frac{27}{x-3}-\frac{x^3}{x-3}=}$${\displaystyle \frac{27-x^3}{x-3}=}$${\displaystyle \frac{-\cancel{\left(x-3\right)}(x^2+3x+9)}{\cancel{x-3}}=}$${\displaystyle -x^2-3x-9}$;

5) ${\displaystyle \frac{8}{a-2}-\frac{a^3}{a-2}=}$${\displaystyle \frac{8-a^3}{a-2}=}$${\displaystyle \frac{-\cancel{\left(a-2\right)}(a^2+2a+4)}{\cancel{a-2}}=}$${\displaystyle -a^2-2a-4}$;

6) ${\displaystyle \frac{a}{a^3+b^3}+\frac{b}{a^3+b^3}=}$${\displaystyle \frac{\cancel{a+b}}{\cancel{\left(a+b\right)}(a^2-ab+b^2)}=}$${\displaystyle \frac{1}{a^2-ab+b^2}}$;

7) ${\displaystyle \frac{c}{b^3-c^3}-\frac{b}{b^3-c^3}=}$${\displaystyle \frac{c-b}{b^3-c^3}=}$${\displaystyle \frac{-\cancel{\left(b-c\right)}}{\cancel{\left(b-c\right)}(b^2+bc+c^2)}=}$${\displaystyle -\frac{1}{b^2+bc+c^2}}$;

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

9) ${\displaystyle \frac{a^2+b^2}{a^3+b^3}-\frac{ab}{a^3+b^3}=}$${\displaystyle \frac{\cancel{a^2-ab+b^2}}{(a+b)\cancel{\left(a^2-ab+b^2\right)}}=}$${\displaystyle \frac{1}{a+b}}$;

10) ${\displaystyle \frac{c^2}{b^3-c^3}+\frac{bc+b^2}{b^3-c^3}=}$${\displaystyle \frac{\cancel{b^2+bc+c^2}}{(b-c)\cancel{\left(b^2+bc+c^2\right)}}=}$${\displaystyle \frac{1}{b-c}}$.