557 Вправа 557Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 162

557. Знайдіть значення виразу:

1) ${\displaystyle \sqrt{14}\cdot \sqrt{\frac{2}{7}}+\sqrt{12,96}}$;

${\displaystyle \sqrt{14}\cdot \sqrt{\frac{2}{7}}=}$${\displaystyle \sqrt{\frac{14\cdot 2}{7}}=}$${\displaystyle \sqrt{2\cdot 2}=}$${\displaystyle \sqrt{4}=}$${\displaystyle 2}$;

${\displaystyle \sqrt{12,96}=}$${\displaystyle \sqrt{\frac{1296}{100}}=}$${\displaystyle \frac{\sqrt{1296}}{\sqrt{100}}=}$${\displaystyle \frac{36}{10}=}$${\displaystyle 3,6}$;

${\displaystyle 2+3,6=}$${\displaystyle 5,6}$;

2) ${\displaystyle \sqrt{0}\cdot \sqrt{2\frac{7}{9}}-\sqrt{1,44}=}$${\displaystyle 0\cdot \sqrt{2\frac{7}{9}}-1,2=}$${\displaystyle 0-1,2=}$${\displaystyle -1,2}$;

3) ${\displaystyle \sqrt{0}+\sqrt{3\frac{1}{2}}\cdot \sqrt{56}=}$${\displaystyle 0+\sqrt{\frac{7}{2}}\cdot \sqrt{56}=}$${\displaystyle \sqrt{\frac{7\cdot 56}{2}}=}$${\displaystyle \sqrt{7\cdot 28}=}$${\displaystyle \sqrt{7\cdot 7\cdot 4}=}$${\displaystyle \sqrt{49\cdot 4}=}$${\displaystyle 7\cdot 2=}$${\displaystyle 14}$;

4) ${\displaystyle 2\frac{1}{11}=}$${\displaystyle \frac{23}{11}}$; ${\displaystyle 1\frac{2}{9}=}$${\displaystyle \frac{11}{9}}$;

${\displaystyle \sqrt{2\frac{1}{11}}\cdot \sqrt{23}\cdot \sqrt{1\frac{2}{9}}=}$${\displaystyle \sqrt{\frac{23}{11}\cdot 23\cdot \frac{11}{9}}=}$${\displaystyle \sqrt{\frac{23\cdot 23\cdot 11}{11\cdot 9}}=}$${\displaystyle \sqrt{\frac{{23}^2}{9}}=}$${\displaystyle \frac{23}{3}=}$${\displaystyle 7\frac{2}{3}}$;

5) ${\displaystyle 3\frac{3}{5}=}$${\displaystyle \frac{18}{5}}$; ${\displaystyle \sqrt{0,01}=}$${\displaystyle 0,1=}$${\displaystyle \frac{1}{10}}$;

${\displaystyle \sqrt{10}:\sqrt{3\frac{3}{5}}\cdot \sqrt{0,01}=}$${\displaystyle \sqrt{10:\frac{18}{5}}\cdot \frac{1}{10}=}$${\displaystyle \sqrt{\frac{10\cdot 5}{18}}\cdot \frac{1}{10}=}$${\displaystyle \sqrt{\frac{25}{9}}\cdot \frac{1}{10}=}$${\displaystyle \frac{5}{3}\cdot \frac{1}{10}=}$${\displaystyle \frac{1}{6}}$;

6) ${\displaystyle \frac{\sqrt{48}}{\sqrt{3}}=}$${\displaystyle \sqrt{\frac{48}{3}}=}$${\displaystyle \sqrt{16}=}$${\displaystyle 4}$;

${\displaystyle \frac{\sqrt{14}\cdot \sqrt{35}}{\sqrt{10}}=}$${\displaystyle \sqrt{\frac{14\cdot 35}{10}}=}$${\displaystyle \sqrt{\frac{2\cdot 7\cdot 5\cdot 7}{2\cdot 5}}=}$${\displaystyle \sqrt{7^2}=}$${\displaystyle 7}$;

${\displaystyle 4+7=}$${\displaystyle 11}$.