Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 190

\[\boxed{\mathbf{190}.}\]

\[b_{1} = x;\ \ b_{2} = y;\ \ b_{3} = z;\]

\[a_{1} = x;\ \ a_{2} = 2y;\ \ a_{3} = 3z.\]

\[\left\{ \begin{matrix} x = 2y + d\ \\ 2y = 3z + d \\ \end{matrix} \right.\ ( - )\]

\[x - 2y = 2y - 3z\]

\[x - 4y + 3z = 0\ \ \ \ |\ :x\]

\[1 - 4 \cdot \frac{y}{x} + 3 \cdot \frac{z}{x} = 0\]

\[\left\{ \begin{matrix} q = \frac{y}{x} \\ q^{2} = \frac{z}{x} \\ \end{matrix}\text{\ \ } \right.\ \]

\[1 - 4q + 3q^{2} = 0\]

\[3q^{2} - 4q + 1 = 0\]

\[D_{1} = 4 - 3 = 1\]

\[q_{1} = \frac{2 + 1}{3} = 1;\ \]

\[\ q_{2} = \frac{2 - 1}{3} = \frac{1}{3}.\]

\[Ответ:1\ или\ \frac{1}{3}.\]