536° Вправа 536Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 158

536. Внесіть множник під знак кореня:

1) ${\displaystyle 2\sqrt{2}=}$${\displaystyle \sqrt{2^2\cdot 2}=}$${\displaystyle \sqrt{4\cdot 2}=}$${\displaystyle \sqrt{8}}$;

2) ${\displaystyle 4\sqrt{2}=}$${\displaystyle \sqrt{4^2\cdot 2}=}$${\displaystyle \sqrt{16\cdot 2}=}$${\displaystyle \sqrt{32}}$;

3) ${\displaystyle 2\sqrt{3}=}$${\displaystyle \sqrt{2^2\cdot 3}=}$${\displaystyle \sqrt{4\cdot 3}=}$${\displaystyle \sqrt{12}}$;

4) ${\displaystyle 5\sqrt{3}=}$${\displaystyle \sqrt{5^2\cdot 3}=}$${\displaystyle \sqrt{25\cdot 3}=}$${\displaystyle \sqrt{75}}$;

5) ${\displaystyle 3\sqrt{5}=}$${\displaystyle \sqrt{3^2\cdot 5}=}$${\displaystyle \sqrt{9\cdot 5}=}$${\displaystyle \sqrt{45}}$;

6) ${\displaystyle 5\sqrt{6}=}$${\displaystyle \sqrt{5^2\cdot 6}=}$${\displaystyle \sqrt{25\cdot 6}=}$${\displaystyle \sqrt{150}}$;

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

8) ${\displaystyle 3\sqrt{7}=}$${\displaystyle \sqrt{3^2\cdot 7}=}$${\displaystyle \sqrt{9\cdot 7}=}$${\displaystyle \sqrt{63}}$;

9) ${\displaystyle 2\sqrt{10}=}$${\displaystyle \sqrt{2^2\cdot 10}=}$${\displaystyle \sqrt{4\cdot 10}=}$${\displaystyle \sqrt{40}}$;

10) ${\displaystyle 3\sqrt{10}=}$${\displaystyle \sqrt{3^2\cdot 10}=}$${\displaystyle \sqrt{9\cdot 10}=}$${\displaystyle \sqrt{90}}$;

11) ${\displaystyle 3\sqrt{11}=}$${\displaystyle \sqrt{3^2\cdot 11}=}$${\displaystyle \sqrt{9\cdot 11}=}$${\displaystyle \sqrt{99}}$;

12) ${\displaystyle 2\sqrt{13}=}$${\displaystyle \sqrt{2^2\cdot 13}=}$${\displaystyle \sqrt{4\cdot 13}=}$${\displaystyle \sqrt{52}}$.