568 Вправа 568Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 163

568. Розкладіть на множники:

1) ${\displaystyle \sqrt{30}-\sqrt{3}=}$${\displaystyle \sqrt{3}\cdot \sqrt{10}-\sqrt{3}=}$${\displaystyle \sqrt{3}\left(\sqrt{10}-1\right)}$;

2) ${\displaystyle \sqrt{17}+3\sqrt{34}=}$${\displaystyle \sqrt{17}+3\sqrt{17}\cdot \sqrt{2}=}$${\displaystyle \sqrt{17}\left(1+3\sqrt{2}\right)}$;

3) ${\displaystyle \sqrt{5}+5=}$${\displaystyle \sqrt{5}+\sqrt{5}\cdot \sqrt{5}=}$${\displaystyle \sqrt{5}\left(1+\sqrt{5}\right)}$;

4) ${\displaystyle 10-\sqrt{10}=}$${\displaystyle \sqrt{10}\cdot \sqrt{10}-\sqrt{10}=}$${\displaystyle \sqrt{10}\left(\sqrt{10}-1\right)}$;

5) ${\displaystyle \sqrt{3}+6=}$${\displaystyle \sqrt{3}+2\sqrt{3}\cdot \sqrt{3}=}$${\displaystyle \sqrt{3}\left(1+2\sqrt{3}\right)}$;

6) ${\displaystyle \sqrt{12}-\sqrt{15}+3\sqrt{42}=}$${\displaystyle 2\sqrt{3}-\sqrt{3}\cdot \sqrt{5}+3\sqrt{3}\cdot \sqrt{14}=}$${\displaystyle \sqrt{3}\left(2-\sqrt{5}+3\sqrt{14}\right)}$.