737 Вправа 737Алгебра 8 клас Бевз 2025 · § 15 · стор. 143

737. Обчисли значення добутку.

а) ${\displaystyle \sqrt{\frac{2}{3}}\cdot \sqrt{\frac{6}{25}}=}$${\displaystyle \sqrt{\frac{2\cdot 6}{3\cdot 25}}=}$${\displaystyle \sqrt{\frac{2\cdot 2\cdot \cancel{3}}{\cancel{3}\cdot 25}}=}$${\displaystyle \sqrt{\frac{4}{25}}=}$${\displaystyle \frac{2}{5}}$;

б) ${\displaystyle \sqrt{1\frac{1}{16}}\cdot \sqrt{1\frac{8}{17}}=}$${\displaystyle \sqrt{\frac{17}{16}\cdot \frac{25}{17}}=}$${\displaystyle \sqrt{\frac{\cancel{17}\cdot 25}{16\cdot \cancel{17}}}=}$${\displaystyle \sqrt{\frac{25}{16}}}$=

=${\displaystyle \frac{5}{4}=}$${\displaystyle 1\frac{1}{4}}$;

в) ${\displaystyle \sqrt{2\frac{1}{16}}\cdot \sqrt{\frac{3}{11}}=}$${\displaystyle \sqrt{\frac{33}{16}\cdot \frac{3}{11}}=}$${\displaystyle \sqrt{\frac{3\cdot \cancel{11}\cdot 3}{16\cdot \cancel{11}}}=}$${\displaystyle \sqrt{\frac{9}{16}}=}$${\displaystyle \frac{3}{4}}$.