Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 809

\[\boxed{\mathbf{809}.}\]

\[a = \log_{2}3;b = \log_{3}5:\]

\[\log_{300}120 = \frac{lg120}{lg300} =\]

\[= \frac{\lg(3 \cdot 4 \cdot 10)}{lg300} =\]

\[= \frac{lg3 + lg4 + lg10}{lg300} =\]

\[= \frac{lg3 + lg2^{2} + 1}{lg300} =\]

\[= \frac{lg3 + 2lg2 + 1}{\lg(3 \cdot 100)} =\]

\[= \frac{lg3 + 2lg2 + lg1}{lg3 + lg10^{2}} =\]

\[= \frac{lg3 + 2lg2 + 1}{lg3 + 2};\]

\[a = \log_{2}3 = \frac{lg3}{lg2} \rightarrow lg3 = a \cdot lg2;\]

\[b = \log_{3}5 = \log_{3}(10\ :2) \rightarrow b =\]

\[= \log_{3}10 - \log_{3}2 = \frac{1}{lg3} -\]

\[- \frac{lg2}{lg3} = \frac{1}{alg2} - \frac{lg2}{alg2} =\]

\[= \frac{1 - lg2}{alg2}\]

\[b \cdot alg2 = 1 - lg2\]

\[\text{ab}lg2 + lg2 = 1\]

\[lg2 \cdot (ab + 1) = 1\]

\[lg2 = \frac{1}{ab + 1}.\]

\[\frac{lg3 + 2lg2 + 1}{lg3 + 2} =\]

\[= \frac{a \cdot lg2 + 2lg2 + 1}{alg2 + 2} =\]

\[= \frac{a \cdot \frac{1}{ab + 1} + 2 \cdot \frac{1}{ab + 1} + 1}{a \cdot \frac{1}{ab + 1} + 2} =\]

\[= \frac{a + ab + 3}{ab + 1}\ :\frac{a + 2ab + 2}{ab + 1} =\]

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

\[= \frac{a + ab + 3}{a + 2ab + 2}.\]