Страница 42 номер 2.3.18, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Заполнить таблицу значений алгебраических дробей: \[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{№} & a & b & \frac{a}{b} & \frac{b}{a} & \frac{a + b}{a} & \frac{b}{a - b} & \frac{a - b^2}{a + b^2} & \frac{a^2 - b^2}{2a} \\ \hline 1 & 1 & -2 & & & & & & \\ \hline 2 & 2 & -1 & & & & & & \\ \hline 3 & 0{,}3 & 0{,}4 & & & & & & \\ \hline 4 & -\frac{1}{2} & -\frac{1}{3} & & & & & & \\ \hline \end{array}\]
Знаменатели \(b\), \(a\), \(a - b\), \(a + b^2\) и \(2a\) ни при одной паре значений в нуль не обращаются, поэтому все дроби имеют смысл.
\[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{№} & a & b & \frac{a}{b} & \frac{b}{a} & \frac{a + b}{a} & \frac{b}{a - b} & \frac{a - b^2}{a + b^2} & \frac{a^2 - b^2}{2a} \\ \hline 1 & 1 & -2 & -\frac{1}{2} & -2 & -1 & -\frac{2}{3} & -\frac{3}{5} & -\frac{3}{2} \\ \hline 2 & 2 & -1 & -2 & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{3} & \frac{1}{3} & \frac{3}{4} \\ \hline 3 & 0{,}3 & 0{,}4 & \frac{3}{4} & \frac{4}{3} & \frac{7}{3} & -4 & \frac{7}{23} & -\frac{7}{60} \\ \hline 4 & -\frac{1}{2} & -\frac{1}{3} & \frac{3}{2} & \frac{2}{3} & \frac{5}{3} & 2 & \frac{11}{7} & -\frac{5}{36} \\ \hline \end{array}\]
При \(a = 0{,}3\), \(b = 0{,}4\): \(\frac{a - b^2}{a + b^2} = \frac{0{,}3 - 0{,}16}{0{,}3 + 0{,}16} = \frac{0{,}14}{0{,}46} = \frac{7}{23}\), \(\frac{a^2 - b^2}{2a} = \frac{0{,}09 - 0{,}16}{0{,}6} = \frac{-0{,}07}{0{,}6} = -\frac{7}{60}\).
При \(a = -\frac{1}{2}\), \(b = -\frac{1}{3}\): \(a + b = -\frac{5}{6}\), \(a - b = -\frac{1}{6}\), \(b^2 = \frac{1}{9}\), \(a - b^2 = -\frac{11}{18}\), \(a + b^2 = -\frac{7}{18}\), \(a^2 - b^2 = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}\), \(2a = -1\).
Ответ: при \(a = 1\), \(b = -2\): \(\frac{a}{b} = -\frac{1}{2}\), \(\frac{b}{a} = -2\), \(\frac{a + b}{a} = -1\), \(\frac{b}{a - b} = -\frac{2}{3}\), \(\frac{a - b^2}{a + b^2} = -\frac{3}{5}\), \(\frac{a^2 - b^2}{2a} = -\frac{3}{2}\); при \(a = 2\), \(b = -1\): \(\frac{a}{b} = -2\), \(\frac{b}{a} = -\frac{1}{2}\), \(\frac{a + b}{a} = \frac{1}{2}\), \(\frac{b}{a - b} = -\frac{1}{3}\), \(\frac{a - b^2}{a + b^2} = \frac{1}{3}\), \(\frac{a^2 - b^2}{2a} = \frac{3}{4}\); при \(a = 0{,}3\), \(b = 0{,}4\): \(\frac{a}{b} = \frac{3}{4}\), \(\frac{b}{a} = \frac{4}{3}\), \(\frac{a + b}{a} = \frac{7}{3}\), \(\frac{b}{a - b} = -4\), \(\frac{a - b^2}{a + b^2} = \frac{7}{23}\), \(\frac{a^2 - b^2}{2a} = -\frac{7}{60}\); при \(a = -\frac{1}{2}\), \(b = -\frac{1}{3}\): \(\frac{a}{b} = \frac{3}{2}\), \(\frac{b}{a} = \frac{2}{3}\), \(\frac{a + b}{a} = \frac{5}{3}\), \(\frac{b}{a - b} = 2\), \(\frac{a - b^2}{a + b^2} = \frac{11}{7}\), \(\frac{a^2 - b^2}{2a} = -\frac{5}{36}\).
Заполнить таблицу значений алгебраических дробей: \[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{№} & a & b & \frac{a}{b} & \frac{b}{a} & \frac{a + b}{a} & \frac{b}{a - b} & \frac{a - b^2}{a + b^2} & \frac{a^2 - b^2}{2a} \\ \hline 1 & 1 & -2 & & & & & & \\ \hline 2 & 2 & -1 & & & & & & \\ \hline 3 & 0{,}3 & 0{,}4 & & & & & & \\ \hline 4 & -\frac{1}{2} & -\frac{1}{3} & & & & & & \\ \hline \end{array}\]
Знаменатели данных дробей — \(b\), \(a\), \(a - b\), \(a + b^2\) и \(2a\); ни при одной паре значений из таблицы они в нуль не обращаются, поэтому все дроби имеют смысл.
\[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{№} & a & b & \frac{a}{b} & \frac{b}{a} & \frac{a + b}{a} & \frac{b}{a - b} & \frac{a - b^2}{a + b^2} & \frac{a^2 - b^2}{2a} \\ \hline 1 & 1 & -2 & -\frac{1}{2} & -2 & -1 & -\frac{2}{3} & -\frac{3}{5} & -\frac{3}{2} \\ \hline 2 & 2 & -1 & -2 & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{3} & \frac{1}{3} & \frac{3}{4} \\ \hline 3 & 0{,}3 & 0{,}4 & \frac{3}{4} & \frac{4}{3} & \frac{7}{3} & -4 & \frac{7}{23} & -\frac{7}{60} \\ \hline 4 & -\frac{1}{2} & -\frac{1}{3} & \frac{3}{2} & \frac{2}{3} & \frac{5}{3} & 2 & \frac{11}{7} & -\frac{5}{36} \\ \hline \end{array}\]
1) При \(a = 1\), \(b = -2\): \(\frac{a}{b} = \frac{1}{-2} = -\frac{1}{2}\); \(\frac{b}{a} = \frac{-2}{1} = -2\); \(\frac{a + b}{a} = \frac{1 - 2}{1} = -1\); \(\frac{b}{a - b} = \frac{-2}{1 - (-2)} = \frac{-2}{3} = -\frac{2}{3}\); \(\frac{a - b^2}{a + b^2} = \frac{1 - 4}{1 + 4} = -\frac{3}{5}\); \(\frac{a^2 - b^2}{2a} = \frac{1 - 4}{2} = -\frac{3}{2}\).
2) При \(a = 2\), \(b = -1\): \(\frac{a}{b} = \frac{2}{-1} = -2\); \(\frac{b}{a} = \frac{-1}{2} = -\frac{1}{2}\); \(\frac{a + b}{a} = \frac{2 - 1}{2} = \frac{1}{2}\); \(\frac{b}{a - b} = \frac{-1}{2 - (-1)} = \frac{-1}{3} = -\frac{1}{3}\); \(\frac{a - b^2}{a + b^2} = \frac{2 - 1}{2 + 1} = \frac{1}{3}\); \(\frac{a^2 - b^2}{2a} = \frac{4 - 1}{4} = \frac{3}{4}\).
3) При \(a = 0{,}3\), \(b = 0{,}4\): \(\frac{a}{b} = \frac{0{,}3}{0{,}4} = \frac{3}{4}\); \(\frac{b}{a} = \frac{0{,}4}{0{,}3} = \frac{4}{3}\); \(\frac{a + b}{a} = \frac{0{,}7}{0{,}3} = \frac{7}{3}\); \(\frac{b}{a - b} = \frac{0{,}4}{0{,}3 - 0{,}4} = \frac{0{,}4}{-0{,}1} = -4\); \(\frac{a - b^2}{a + b^2} = \frac{0{,}3 - 0{,}16}{0{,}3 + 0{,}16} = \frac{0{,}14}{0{,}46} = \frac{7}{23}\); \(\frac{a^2 - b^2}{2a} = \frac{0{,}09 - 0{,}16}{0{,}6} = \frac{-0{,}07}{0{,}6} = -\frac{7}{60}\).
4) При \(a = -\frac{1}{2}\), \(b = -\frac{1}{3}\): \(a + b = -\frac{5}{6}\), \(a - b = -\frac{1}{6}\), \(b^2 = \frac{1}{9}\), \(a - b^2 = -\frac{11}{18}\), \(a + b^2 = -\frac{7}{18}\), \(a^2 - b^2 = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}\), \(2a = -1\). Тогда \(\frac{a}{b} = \left(-\frac{1}{2}\right) : \left(-\frac{1}{3}\right) = \frac{3}{2}\); \(\frac{b}{a} = \left(-\frac{1}{3}\right) : \left(-\frac{1}{2}\right) = \frac{2}{3}\); \(\frac{a + b}{a} = \left(-\frac{5}{6}\right) : \left(-\frac{1}{2}\right) = \frac{5}{3}\); \(\frac{b}{a - b} = \left(-\frac{1}{3}\right) : \left(-\frac{1}{6}\right) = 2\); \(\frac{a - b^2}{a + b^2} = \left(-\frac{11}{18}\right) : \left(-\frac{7}{18}\right) = \frac{11}{7}\); \(\frac{a^2 - b^2}{2a} = \frac{5}{36} : (-1) = -\frac{5}{36}\).
Ответ: при \(a = 1\), \(b = -2\): \(\frac{a}{b} = -\frac{1}{2}\), \(\frac{b}{a} = -2\), \(\frac{a + b}{a} = -1\), \(\frac{b}{a - b} = -\frac{2}{3}\), \(\frac{a - b^2}{a + b^2} = -\frac{3}{5}\), \(\frac{a^2 - b^2}{2a} = -\frac{3}{2}\); при \(a = 2\), \(b = -1\): \(\frac{a}{b} = -2\), \(\frac{b}{a} = -\frac{1}{2}\), \(\frac{a + b}{a} = \frac{1}{2}\), \(\frac{b}{a - b} = -\frac{1}{3}\), \(\frac{a - b^2}{a + b^2} = \frac{1}{3}\), \(\frac{a^2 - b^2}{2a} = \frac{3}{4}\); при \(a = 0{,}3\), \(b = 0{,}4\): \(\frac{a}{b} = \frac{3}{4}\), \(\frac{b}{a} = \frac{4}{3}\), \(\frac{a + b}{a} = \frac{7}{3}\), \(\frac{b}{a - b} = -4\), \(\frac{a - b^2}{a + b^2} = \frac{7}{23}\), \(\frac{a^2 - b^2}{2a} = -\frac{7}{60}\); при \(a = -\frac{1}{2}\), \(b = -\frac{1}{3}\): \(\frac{a}{b} = \frac{3}{2}\), \(\frac{b}{a} = \frac{2}{3}\), \(\frac{a + b}{a} = \frac{5}{3}\), \(\frac{b}{a - b} = 2\), \(\frac{a - b^2}{a + b^2} = \frac{11}{7}\), \(\frac{a^2 - b^2}{2a} = -\frac{5}{36}\).
