Страница 204 номер 583, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(\frac{a + 1}{a - 1} + \frac{6}{a^{2} - 1} - \frac{a + 3}{a + 1}\) при \(a = -9\)
\(\dfrac{a + 1}{a - 1} + \dfrac{6}{a^{2} - 1} - \dfrac{a + 3}{a + 1} = \dfrac{(a + 1)^{2} + 6 - (a + 3)(a - 1)}{a^{2} - 1} =\)
\(= \dfrac{a^{2} + 2a + 1 + 6 - a^{2} - 2a + 3}{a^{2} - 1} = \dfrac{10}{a^{2} - 1},\)
\(\dfrac{10}{a^{2} - 1} = \dfrac{10}{81 - 1} = \dfrac{1}{8};\)

2) \(\frac{b + 5}{b + 2} - \frac{3}{b^{2} - 4} - \frac{b + 1}{b - 2}\) при \(b = -8\)
\(\dfrac{b + 5}{b + 2} - \dfrac{3}{b^{2} - 4} - \dfrac{b + 1}{b - 2} = \dfrac{(b + 5)(b - 2) - 3 - (b + 1)(b + 2)}{b^{2} - 4} =\)
\(= \dfrac{b^{2} - 2b + 5b - 10 - 3 - b^{2} - 3b - 2}{b^{2} - 4} =\)
\(= \dfrac{-15}{b^{2} - 4},\)
\(\dfrac{-15}{b^{2} - 4} = -\dfrac{15}{64 - 4} = -\dfrac{1}{4};\)

3) \(\frac{a - 2}{a - 3} : \left(\frac{a^{2} - 6a + 10}{a^{2} - 9} + \frac{2}{a + 3}\right)\) при \(a = -1\frac{1}{2}\)
\(\dfrac{a - 2}{a - 3} : \left(\dfrac{a^{2} - 6a + 10}{a^{2} - 9} + \dfrac{2}{a + 3}\right) =\)
\(= \dfrac{a - 2}{a - 3} : \dfrac{a^{2} - 6a + 10 + 2a - 6}{a^{2} - 9} = \dfrac{a - 2}{a - 3} \cdot \dfrac{(a - 3)(a + 3)}{a^{2} - 4a + 4} =\)
\(= \dfrac{(a - 2)(a + 3)}{(a - 2)^{2}} = \dfrac{a + 3}{a - 2},\)
\(\dfrac{a + 3}{a - 2} = \dfrac{-\frac{3}{2} + 3}{-\dfrac{3}{2} - 2} = \dfrac{3}{2} \cdot \left(-\dfrac{2}{7}\right) = -\dfrac{3}{7};\)

4) \(\frac{b + 1}{b - 4} : \left(\frac{b^{2} + 9}{b^{2} - 16} + \frac{2}{b + 4}\right)\) при \(b = 4\frac{1}{3}\)
\(\dfrac{b + 1}{b - 4} : \left(\dfrac{b^{2} + 9}{b^{2} - 16} + \dfrac{2}{b + 4}\right) = \dfrac{b + 1}{b - 4} : \dfrac{b^{2} + 9 + 2b - 8}{(b - 4)(b + 4)} =\)
\(= \dfrac{b + 1}{b - 4} \cdot \dfrac{(b - 4)(b + 4)}{b^{2} + 2b + 1} = \dfrac{b + 4}{b + 1},\)
\(\dfrac{b + 4}{b + 1} = \dfrac{\frac{13}{3} + 4}{\dfrac{13}{3} + 1} = \dfrac{25}{3} \cdot \dfrac{3}{16} = \dfrac{25}{16} = 1\dfrac{9}{16}.\)