758 Вправа 758Алгебра 8 клас Бевз 2025 · § 15 · стор. 145

758. Обчисли значення виразу.

а) ${\displaystyle \sqrt{4\frac{3}{8}}\cdot \sqrt{1\frac{23}{40}}=}$${\displaystyle \sqrt{\frac{35}{8}\cdot \frac{63}{40}}=}$${\displaystyle \sqrt{\frac{7\cdot \cancel{5}\cdot 63}{8\cdot 8\cdot \cancel{5}}}=}$${\displaystyle \sqrt{\frac{441}{64}}}$=

=${\displaystyle \frac{21}{8}=}$${\displaystyle 2\frac{5}{8}}$;

б) ${\displaystyle \sqrt{3\frac{5}{7}}\cdot \sqrt{13}\cdot \sqrt{\frac{9}{14}}=}$${\displaystyle \sqrt{\frac{26}{7}\cdot 13\cdot \frac{9}{14}}}$=

=${\displaystyle \sqrt{\frac{\cancel{2}\cdot 13\cdot 13\cdot 9}{7\cdot \cancel{2}\cdot 7}}=}$${\displaystyle \sqrt{\frac{1521}{49}}=}$${\displaystyle \frac{39}{7}=}$${\displaystyle 5\frac{4}{7}}$;

в) ${\displaystyle \sqrt{10\frac{5}{7}}\cdot \sqrt{\frac{8}{11}}\cdot \sqrt{1\frac{1}{21}}=}$${\displaystyle \sqrt{\frac{75}{7}\cdot \frac{8}{11}\cdot \frac{22}{21}}}$=

=${\displaystyle \sqrt{\frac{25\cdot \cancel{3}\cdot 8\cdot 2\cdot \cancel{11}}{7\cdot \cancel{11}\cdot 7\cdot \cancel{3}}}=}$${\displaystyle \sqrt{\frac{400}{49}}=}$${\displaystyle \frac{20}{7}=}$${\displaystyle 2\frac{6}{7}}$.