539° Вправа 539Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 159

539. Винесіть множник з-під знака кореня:

1) ${\displaystyle \sqrt{18}=}$${\displaystyle \sqrt{9\cdot 2}=}$${\displaystyle \sqrt{9}\cdot \sqrt{2}=}$${\displaystyle 3\sqrt{2}}$;

2) ${\displaystyle \sqrt{50}=}$${\displaystyle \sqrt{25\cdot 2}=}$${\displaystyle \sqrt{25}\cdot \sqrt{2}=}$${\displaystyle 5\sqrt{2}}$;

3) ${\displaystyle \sqrt{98}=}$${\displaystyle \sqrt{49\cdot 2}=}$${\displaystyle \sqrt{49}\cdot \sqrt{2}=}$${\displaystyle 7\sqrt{2}}$;

4) ${\displaystyle \sqrt{28}=}$${\displaystyle \sqrt{4\cdot 7}=}$${\displaystyle \sqrt{4}\cdot \sqrt{7}=}$${\displaystyle 2\sqrt{7}}$;

5) ${\displaystyle \sqrt{200}=}$${\displaystyle \sqrt{100\cdot 2}=}$${\displaystyle \sqrt{100}\cdot \sqrt{2}=}$${\displaystyle 10\sqrt{2}}$;

6) ${\displaystyle \sqrt{27}=}$${\displaystyle \sqrt{9\cdot 3}=}$${\displaystyle \sqrt{9}\cdot \sqrt{3}=}$${\displaystyle 3\sqrt{3}}$;

7) ${\displaystyle \sqrt{48}=}$${\displaystyle \sqrt{16\cdot 3}=}$${\displaystyle \sqrt{16}\cdot \sqrt{3}=}$${\displaystyle 4\sqrt{3}}$;

8) ${\displaystyle \sqrt{108}=}$${\displaystyle \sqrt{36\cdot 3}=}$${\displaystyle \sqrt{36}\cdot \sqrt{3}=}$${\displaystyle 6\sqrt{3}}$.