641° Вправа 641Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 191

641. Звільніться від ірраціональності в знаменнику дробу:

1) ${\displaystyle \frac{1^{\mathrm{\setminus }\sqrt{5}}}{\sqrt{5}}=}$${\displaystyle \frac{\sqrt{5}}{5}}$;

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

3) ${\displaystyle \frac{4^{\mathrm{\setminus }\sqrt{5}}}{3\sqrt{5}}=}$${\displaystyle \frac{4\sqrt{5}}{3\cdot 5}=}$${\displaystyle \frac{4\sqrt{5}}{15}}$;

4) ${\displaystyle \frac{{10}^{\mathrm{\setminus }\sqrt{10}}}{\sqrt{10}}=}$${\displaystyle \frac{10\sqrt{10}}{10}=}$${\displaystyle \sqrt{10}}$;

5) ${\displaystyle \frac{{14}^{\mathrm{\setminus }\sqrt{7}}}{\sqrt{7}}=}$${\displaystyle \frac{14\sqrt{7}}{7}=}$${\displaystyle 2\sqrt{7}}$;

6) ${\displaystyle \frac{2^{\mathrm{\setminus }\sqrt{11}}}{\sqrt{11}}=}$${\displaystyle \frac{2\sqrt{11}}{11}}$;

7) ${\displaystyle \frac{1^{\mathrm{\setminus }\sqrt{m}}}{\sqrt{m}}=}$${\displaystyle \frac{\sqrt{m}}{m}}$;

8) ${\displaystyle \frac{2^{\mathrm{\setminus }\sqrt{n}}}{\sqrt{n}}=}$${\displaystyle \frac{2\sqrt{n}}{n}}$;

9) ${\displaystyle \frac{y^{\mathrm{\setminus }\sqrt{y}}}{\sqrt{y}}=}$${\displaystyle \frac{y\sqrt{y}}{y}=}$${\displaystyle \sqrt{y}}$;

10) ${\displaystyle \frac{1^{\mathrm{\setminus }\sqrt{mn}}}{\sqrt{mn}}=}$${\displaystyle \frac{\sqrt{mn}}{mn}}$;

11) ${\displaystyle \frac{1^{\mathrm{\setminus }\sqrt{x-1}}}{\sqrt{x-1}}=}$${\displaystyle \frac{\sqrt{x-1}}{x-1}}$;

12) ${\displaystyle \frac{1^{\mathrm{\setminus }\sqrt{m+5}}}{\sqrt{m+5}}=}$${\displaystyle \frac{\sqrt{m+5}}{m+5}}$.