160 Вправа 160Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 49

160. Виконайте дії.

1) ${\displaystyle \frac{y-2}{5}+\frac{3+y}{5}+\frac{1-y}{5}=}$${\displaystyle \frac{y-2+3+y+1-y}{5}=}$${\displaystyle \frac{y+2}{5}}$;

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

3) ${\displaystyle \frac{64}{a-4}-\frac{a^3}{a-4}=}$${\displaystyle \frac{64-a^3}{a-4}=}$${\displaystyle \frac{-\cancel{\left(a-4\right)}(a^2+4a+16)}{\cancel{a-4}}=}$${\displaystyle -\left(a^2+4a+16\right)}$;

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

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

6) ${\displaystyle \frac{y^2}{x^3-y^3}+\frac{xy+x^2}{x^3-y^3}=}$${\displaystyle \frac{\cancel{x^2+xy+y^2}}{(x-y)\cancel{\left(x^2+xy+y^2\right)}}=}$${\displaystyle \frac{1}{x-y}}$.