Страница 47 номер 2.4.1, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Заполнить таблицу: \[\begin{array}{|c|c|c|c|c|c|c|c|} \hline № & a & \sqrt{2} & \sqrt{3} & 2\sqrt{3} & 3\sqrt{2} & 2\sqrt{2} & 3\sqrt{3} \\ \hline 1 & a^2 & & & & & & \\ \hline 2 & \frac{a^2}{2} & & & & & & \\ \hline 3 & a^3 & & & & & & \\ \hline 4 & \frac{1}{6}a^4 & & & & & & \\ \hline \end{array}\]
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline № & a & \sqrt{2} & \sqrt{3} & 2\sqrt{3} & 3\sqrt{2} & 2\sqrt{2} & 3\sqrt{3} \\ \hline 1 & a^2 & 2 & 3 & 12 & 18 & 8 & 27 \\ \hline 2 & \frac{a^2}{2} & 1 & 1\frac{1}{2} & 6 & 9 & 4 & 13\frac{1}{2} \\ \hline 3 & a^3 & 2\sqrt{2} & 3\sqrt{3} & 24\sqrt{3} & 54\sqrt{2} & 16\sqrt{2} & 81\sqrt{3} \\ \hline 4 & \frac{1}{6}a^4 & \frac{2}{3} & 1\frac{1}{2} & 24 & 54 & 10\frac{2}{3} & 121\frac{1}{2} \\ \hline \end{array}\]
1) \(a^2\): \(\left(\sqrt{2}\right)^2 = 2\); \(\left(\sqrt{3}\right)^2 = 3\); \(\left(2\sqrt{3}\right)^2 = 2^2 \cdot 3 = 12\); \(\left(3\sqrt{2}\right)^2 = 3^2 \cdot 2 = 18\); \(\left(2\sqrt{2}\right)^2 = 2^2 \cdot 2 = 8\); \(\left(3\sqrt{3}\right)^2 = 3^2 \cdot 3 = 27\);
2) \(\frac{a^2}{2}\): \(\frac{2}{2} = 1\); \(\frac{3}{2} = 1\frac{1}{2}\); \(\frac{12}{2} = 6\); \(\frac{18}{2} = 9\); \(\frac{8}{2} = 4\); \(\frac{27}{2} = 13\frac{1}{2}\);
3) \(a^3 = a^2 \cdot a\): \(2 \cdot \sqrt{2} = 2\sqrt{2}\); \(3 \cdot \sqrt{3} = 3\sqrt{3}\); \(12 \cdot 2\sqrt{3} = 24\sqrt{3}\); \(18 \cdot 3\sqrt{2} = 54\sqrt{2}\); \(8 \cdot 2\sqrt{2} = 16\sqrt{2}\); \(27 \cdot 3\sqrt{3} = 81\sqrt{3}\);
4) \(\frac{1}{6}a^4 = \frac{1}{6}\left(a^2\right)^2\): \(\frac{2^2}{6} = \frac{4}{6} = \frac{2}{3}\); \(\frac{3^2}{6} = \frac{9}{6} = 1\frac{1}{2}\); \(\frac{12^2}{6} = \frac{144}{6} = 24\); \(\frac{18^2}{6} = \frac{324}{6} = 54\); \(\frac{8^2}{6} = \frac{64}{6} = 10\frac{2}{3}\); \(\frac{27^2}{6} = \frac{729}{6} = 121\frac{1}{2}\).
Ответ: для \(a\), равного \(\sqrt{2}\); \(\sqrt{3}\); \(2\sqrt{3}\); \(3\sqrt{2}\); \(2\sqrt{2}\); \(3\sqrt{3}\), получаются значения: 1) \(a^2\) — 2; 3; 12; 18; 8; 27; 2) \(\frac{a^2}{2}\) — 1; \(1\frac{1}{2}\); 6; 9; 4; \(13\frac{1}{2}\); 3) \(a^3\) — \(2\sqrt{2}\); \(3\sqrt{3}\); \(24\sqrt{3}\); \(54\sqrt{2}\); \(16\sqrt{2}\); \(81\sqrt{3}\); 4) \(\frac{1}{6}a^4\) — \(\frac{2}{3}\); \(1\frac{1}{2}\); 24; 54; \(10\frac{2}{3}\); \(121\frac{1}{2}\).
Заполнить таблицу: \[\begin{array}{|c|c|c|c|c|c|c|c|} \hline № & a & \sqrt{2} & \sqrt{3} & 2\sqrt{3} & 3\sqrt{2} & 2\sqrt{2} & 3\sqrt{3} \\ \hline 1 & a^2 & & & & & & \\ \hline 2 & \frac{a^2}{2} & & & & & & \\ \hline 3 & a^3 & & & & & & \\ \hline 4 & \frac{1}{6}a^4 & & & & & & \\ \hline \end{array}\]
Квадрат числа вида \(k\sqrt{m}\) равен \(k^2 m\), так как \(\left(\sqrt{m}\right)^2 = m\); куб находится как \(a^3 = a^2 \cdot a\), четвёртая степень — как \(a^4 = \left(a^2\right)^2\).
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline № & a & \sqrt{2} & \sqrt{3} & 2\sqrt{3} & 3\sqrt{2} & 2\sqrt{2} & 3\sqrt{3} \\ \hline 1 & a^2 & 2 & 3 & 12 & 18 & 8 & 27 \\ \hline 2 & \frac{a^2}{2} & 1 & 1\frac{1}{2} & 6 & 9 & 4 & 13\frac{1}{2} \\ \hline 3 & a^3 & 2\sqrt{2} & 3\sqrt{3} & 24\sqrt{3} & 54\sqrt{2} & 16\sqrt{2} & 81\sqrt{3} \\ \hline 4 & \frac{1}{6}a^4 & \frac{2}{3} & 1\frac{1}{2} & 24 & 54 & 10\frac{2}{3} & 121\frac{1}{2} \\ \hline \end{array}\]
1) \(a^2\): \(\left(\sqrt{2}\right)^2 = 2\); \(\left(\sqrt{3}\right)^2 = 3\); \(\left(2\sqrt{3}\right)^2 = 2^2 \cdot 3 = 12\); \(\left(3\sqrt{2}\right)^2 = 3^2 \cdot 2 = 18\); \(\left(2\sqrt{2}\right)^2 = 2^2 \cdot 2 = 8\); \(\left(3\sqrt{3}\right)^2 = 3^2 \cdot 3 = 27\);
2) \(\frac{a^2}{2}\) — каждый найденный квадрат делится на 2: \(\frac{2}{2} = 1\); \(\frac{3}{2} = 1\frac{1}{2}\); \(\frac{12}{2} = 6\); \(\frac{18}{2} = 9\); \(\frac{8}{2} = 4\); \(\frac{27}{2} = 13\frac{1}{2}\);
3) \(a^3 = a^2 \cdot a\): \(2 \cdot \sqrt{2} = 2\sqrt{2}\); \(3 \cdot \sqrt{3} = 3\sqrt{3}\); \(12 \cdot 2\sqrt{3} = 24\sqrt{3}\); \(18 \cdot 3\sqrt{2} = 54\sqrt{2}\); \(8 \cdot 2\sqrt{2} = 16\sqrt{2}\); \(27 \cdot 3\sqrt{3} = 81\sqrt{3}\);
4) \(\frac{1}{6}a^4 = \frac{1}{6}\left(a^2\right)^2\): \(\frac{2^2}{6} = \frac{4}{6} = \frac{2}{3}\); \(\frac{3^2}{6} = \frac{9}{6} = 1\frac{1}{2}\); \(\frac{12^2}{6} = \frac{144}{6} = 24\); \(\frac{18^2}{6} = \frac{324}{6} = 54\); \(\frac{8^2}{6} = \frac{64}{6} = 10\frac{2}{3}\); \(\frac{27^2}{6} = \frac{729}{6} = 121\frac{1}{2}\).
Ответ: для \(a\), равного \(\sqrt{2}\); \(\sqrt{3}\); \(2\sqrt{3}\); \(3\sqrt{2}\); \(2\sqrt{2}\); \(3\sqrt{3}\), получаются значения: 1) \(a^2\) — 2; 3; 12; 18; 8; 27; 2) \(\frac{a^2}{2}\) — 1; \(1\frac{1}{2}\); 6; 9; 4; \(13\frac{1}{2}\); 3) \(a^3\) — \(2\sqrt{2}\); \(3\sqrt{3}\); \(24\sqrt{3}\); \(54\sqrt{2}\); \(16\sqrt{2}\); \(81\sqrt{3}\); 4) \(\frac{1}{6}a^4\) — \(\frac{2}{3}\); \(1\frac{1}{2}\); 24; 54; \(10\frac{2}{3}\); \(121\frac{1}{2}\).
