Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 102

\[\boxed{\mathbf{102}.}\]

\[1)\ \frac{12 + 3}{2} = \frac{15}{2} = 7,5\]

\[\sqrt{12 \cdot 3} = \sqrt{36} = 6\]

\[2)\ \frac{0,6 + 5,4}{2} = \frac{6}{2} = 3\]

\[\sqrt{0,6 \cdot 5,4} = \sqrt{3,24} = 1,8\]

\[3)\ 3\frac{1}{2} = \frac{7}{2} = \frac{28}{8}\]

\[\ \frac{\frac{7}{8} + \frac{28}{8}}{2} = \frac{\frac{35}{8}}{2} = \frac{35}{16}\]

\[\sqrt{\frac{7}{8} \cdot \frac{7}{2}} = \sqrt{\frac{49}{16}} = \frac{7}{4}\]

\[4)\frac{1}{3} = \frac{100}{300};\ \ 0,3 = \frac{3}{100} = \frac{9}{300}.\]

\[\frac{\frac{100}{300} + \frac{9}{300}}{2} = \frac{109}{300}\ :2 = \frac{109}{600}\]

\[\sqrt{\frac{1}{3} \cdot \frac{3}{100}} = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0,1\]

\[\frac{a + b}{2} = \sqrt{\text{ab}}\]

\[\left( \frac{a + b}{2} \right)^{2} = ab\]

\[\frac{a^{2} + 2ab + b^{2}}{4} = ab\ \ \ \ | \cdot 4\]

\[a^{2} + 2ab + b^{2} = 4ab\]

\[a^{2} - 2ab + b^{2} = 0\]

\[(a - b)^{2} = 0\]

\[a - b = 0\]

\[a = b.\]

\[Среднее\ арифметическое\ \]

\[равно\ среднему\ \]

\[геометрическому,\ если\]

\[числа\ a\ и\ \text{b\ }равны.\]