Страница 218 номер 753, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

\(a_{n} = \dfrac{a_{n - 1} + a_{n + 1}}{2} \Rightarrow a_{2} = \dfrac{a_{1} + a_{3}}{2};\ a_{3} = \dfrac{a_{2} + a_{4}}{2}\)
\(2x + 3 = \dfrac{3x + 4 + x^{2}}{2}\)
\(4x + 6 = x^{2} + 3x + 4\)
\(x^{2} - x - 2 = 0\)
\((x + 1)(x - 2) = 0\)
\(x_{1} = - 1;\ x_{2} = 2\)
\(x^{2} = \dfrac{2x + 3 + 2x^{2} + x}{2}\)
\(2x^{2} = 2x^{2} + 3x + 3\)
\(3x = - 3\)
\(x = - 1\)
\(\begin{cases} x_{1} = - 1;\ x_{2} = 2 \\ x = - 1 \end{cases}\ \Rightarrow x = - 1\)
\(a_{1} = 3x + 4 = 3 \cdot ( - 1) + 4 = - 3 + 4 = 1\)
\(a_{2} = 2x + 3 = 2 \cdot ( - 1) + 3 = - 2 + 3 = 1\)
\(a_{3} = x^{2} = ( - 1)^{2} = 1\)
\(a_{4} = 2x^{2} + x = 2 \cdot ( - 1)^{2} + ( - 1) = 2 - 1 = 1\)

Ответ: \(x = - 1\); \(a_{1} = a_{2} = a_{3} = a_{4} = 1\).