Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 399

\[\boxed{\mathbf{399}\mathbf{.}}\]

\[b_{1} + b_{2} + b_{3} = 62\]

\[b_{1} + b_{2} \bullet q + b_{3} \bullet q^{2} = 62\]

\[b_{1} \bullet \left( 1 + q + q^{2} \right) = 62\]

\[Сумма\ десятичных\ \]

\[логарифмов\ равна\ 3:\]

\[\lg b_{1} + \lg b_{2} + \lg b_{3} = 3\]

\[\lg\left( b_{1} \bullet b_{2} \bullet b_{3} \right) = 3\]

\[\lg\left( b_{1} \bullet b_{1} \bullet q \bullet b_{1} \bullet q^{2} \right) = 3\]

\[\lg\left( b_{1}q \right)^{3} = 3\]

\[3\lg\left( b_{1}q \right) = 3\]

\[\lg\left( b_{1}q \right) = 1\]

\[\lg\left( b_{1}q \right) = \lg 10\]

\[b_{1}q = 10\]

\[b_{1} = \frac{10}{q}.\]

\[Подставим\ значение\ b_{1}:\]

\[\frac{10\left( 1 + q + q^{2} \right)}{q} = 62\]

\[\frac{10 + 10q + 10q^{2} - 62q}{q} = 0\]

\[\frac{10q^{2} - 52q + 10}{q} = 0\]

\[10q^{2} - 52q + 10 = 0\]

\[D = 52^{2} - 4 \bullet 10 \bullet 10 =\]

\[= 2704 - 400 = 2304\]

\[q_{1} = \frac{52 - 48}{2 \bullet 10} = \frac{4}{20} = \frac{1}{5} = 0,2;\]

\[q_{2} = \frac{52 + 48}{2 \bullet 10} = \frac{100}{20} = 5.\]

\[1)\ b_{1} = \frac{10}{q} = \frac{10}{0,2} = 50;\]

\[b_{2} = b_{1} \bullet q = 50 \bullet 0,2 = 10;\]

\[b_{3} = b_{2} \bullet q = 10 \bullet 0,2 = 2.\]

\[2)\ b_{1} = \frac{10}{q} = \frac{10}{5} = 2;\]

\[b_{2} = b_{1} \bullet q = 2 \bullet 5 = 10;\]

\[b_{3} = b_{2} \bullet q = 10 \bullet 5 = 50.\]

\[Ответ:\ \ 2;\ 10;\ 50.\]