3. $\sqrt{\frac{16}{25}}+\frac{1}{2}\sqrt{1\frac{7}{9}}-\sqrt{\frac{49}{225}}=$

Question

Вопрос: $Фото задания: 3. $\sqrt{\frac{16}{25}}+\frac{1}{2}\sqrt{1\frac{7}{9}}-\sqrt{\frac{49}{225}}=$$ Фото задания

Accepted Answer

$\sqrt{\frac{16}{25}}=\frac{4}{5}$, $1\frac{7}{9}=\frac{16}{9}$, поэтому $\frac{1}{2}\sqrt{1\frac{7}{9}}=\frac{1}{2}\cdot\frac{4}{3}=\frac{2}{3}$.

$\sqrt{\frac{49}{225}}=\frac{7}{15}$.

\[\frac{4}{5}+\frac{2}{3}-\frac{7}{15}=\frac{12}{15}+\frac{10}{15}-\frac{7}{15}=\frac{15}{15}=1\]

Ответ: 1

3. \(\sqrt{\frac{16}{25}}+\frac{1}{2}\sqrt{1\frac{7}{9}}-\sqrt{\frac{49}{225}}=\)