Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1277

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\[1)\ \frac{a + 2}{a - 2} \bullet \left( \frac{2a^{2} - a - 3}{a^{2} + 5a + 6}\ :\frac{2a - 3}{a - 2} \right)\]

\[2a^{2} - a - 3 = 0\]

\[D = 1 + 24 = 25\]

\[a_{1} = \frac{1 - 5}{2 \bullet 2} = - 1;\ \]

\[a_{2} = \frac{1 + 5}{2 \bullet 2} = \frac{6}{4} = 1,5;\]

\[2(a + 1)(a - 1,5) = 0.\]

\[a^{2} + 5a + 6 = 0\]

\[D = 25 - 24 = 1\]

\[a_{1} = \frac{- 5 - 1}{2} = - 3;\]

\[a_{2} = \frac{- 5 + 1}{2} = - 2;\]

\[(a + 3)(a + 2) = 0.\]

\[\frac{a + 2}{a - 2} \bullet \left( \frac{2(a + 1)(a - 1,5)}{(a + 3)(a + 2)}\ :\frac{2a - 3}{a - 2} \right) =\]

\[= \frac{a + 2}{a - 2} \bullet \frac{(a + 1)(2a - 3)}{(a + 3)(a + 2)} \bullet \frac{a - 2}{2a - 3} =\]

\[= \frac{a + 1}{a + 3}.\]

\[2)\ \left( 2 + \frac{1}{b} \right)\ :\frac{8b^{2} + 8b + 2}{b^{2} - 4b} \bullet \frac{2b + 1}{b} =\]

\[= \frac{2b + 1}{b}\ :\frac{2 \bullet \left( 4b^{2} + 4b + 1 \right)}{b \bullet (b - 4)} \bullet \frac{2b + 1}{b} =\]

\[= \frac{(2b + 1)^{2}}{b^{2}}\ :\frac{2 \bullet (2b + 1)^{2}}{b \bullet (b - 4)} =\]

\[= \frac{(2b + 1)^{2}}{b^{2}} \bullet \frac{b \bullet (b - 4)}{2 \bullet (2b + 1)^{2}} =\]

\[= \frac{b - 4}{2b}.\]