563 Вправа 563Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 163

563. Внесіть множник під знак кореня:

1) ${\displaystyle 0,1\sqrt{3}=}$${\displaystyle \sqrt{{0,1}^2\cdot 3}=}$${\displaystyle \sqrt{0,01\cdot 3}=}$${\displaystyle \sqrt{0,03}}$;

2) ${\displaystyle -3\sqrt{0,1}=}$${\displaystyle -\sqrt{3^2\cdot 0,1}=}$${\displaystyle -\sqrt{9\cdot 0,1}=}$${\displaystyle -\sqrt{0,9}}$;

3) ${\displaystyle 1,1\sqrt{5}=}$${\displaystyle \sqrt{{1,1}^2\cdot 5}=}$${\displaystyle \sqrt{1,21\cdot 5}=}$${\displaystyle \sqrt{6,05}}$;

4) ${\displaystyle -8\sqrt{0,2}=}$${\displaystyle -\sqrt{64\cdot 0,2}=}$${\displaystyle -\sqrt{12,8}}$;

5) ${\displaystyle -2,4\sqrt{2}=}$${\displaystyle -\sqrt{5,76\cdot 2}=}$${\displaystyle -\sqrt{11,52}}$;

6) ${\displaystyle \frac{1}{2}\sqrt{10}=}$${\displaystyle \sqrt{\frac{1}{4}\cdot 10}=}$${\displaystyle \sqrt{2,5}}$;

7) ${\displaystyle \frac{2}{3}\sqrt{2}=}$${\displaystyle \sqrt{\frac{4}{9}\cdot 2}=}$${\displaystyle \sqrt{\frac{8}{9}}}$;

8) ${\displaystyle -\frac{4}{5}\sqrt{0,2}=}$${\displaystyle -\sqrt{\frac{16}{25}\cdot 0,2}=}$${\displaystyle -\sqrt{0,64\cdot 0,2}=}$${\displaystyle -\sqrt{0,128}}$;

9) ${\displaystyle \frac{1}{4}\sqrt{\frac{2}{3}}=}$${\displaystyle \sqrt{\frac{1}{16}\cdot \frac{2}{3}}=}$${\displaystyle \sqrt{\frac{1}{24}}}$;

10) ${\displaystyle \frac{2}{3}\sqrt{\frac{1}{5}}=}$${\displaystyle \sqrt{\frac{4}{9}\cdot \frac{1}{5}}=}$${\displaystyle \sqrt{\frac{4}{45}}}$;

11) ${\displaystyle \frac{1}{3}\sqrt{\frac{1}{3}}=}$${\displaystyle \sqrt{\frac{1}{9}\cdot \frac{1}{3}}=}$${\displaystyle \sqrt{\frac{1}{27}}}$;

12) ${\displaystyle \frac{2}{7}\sqrt{1\frac{1}{6}}=}$${\displaystyle \sqrt{\frac{4}{49}\cdot \frac{7}{6}}=}$${\displaystyle \sqrt{\frac{2}{21}}}$.