Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 329

\[\boxed{\mathbf{329}.}\]

\[\left( x^{2} - 3x + 2 \right)\left( x^{2} - 7x + 12 \right) = 4\]

\[1)\ x^{2} - 3x + 2 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = - 2;\ \ x_{2} = - 3.\]

\[2)\ x^{2} - 7x + 12 = 0\]

\[x_{1} + x_{2} = 7;\ \ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = 3;\ \ x_{2} = 4.\]

\[(x + 3)(x + 2)(x - 3)(x - 4) =\]

\[= 4\]

\[\left( (x - 2)(x - 3) \right)\left( (x - 1)(x - 4) \right) = 4\]

\[\left( x^{2} - 5x + 6 \right)\left( x^{2} - 5x + 4 \right) -\]

\[- 4 = 0\]

\[t = x^{2} - 5x:\]

\[(t + 6)(t + 4) - 4 = 0\]

\[t^{2} + 6t + 4t + 24 - 4 = 0\]

\[t^{2} + 10t + 20 = 0\]

\[D_{1} = 25 - 20 = 5\]

\[t = - 5 \pm \sqrt{5}.\]

\[t = - 5 + \sqrt{5}:\]

\[x^{2} - 5x = - 5 + \sqrt{5}\]

\[x^{2} - 5x + \left( 5 - \sqrt{5} \right) = 0\]

\[D = 25 - 4 \cdot \left( 5 - \sqrt{5} \right) = 25 -\]

\[- 20 + 4\sqrt{5} = 5 + 4\sqrt{5}\]

\[x = \frac{5 \pm \sqrt{5 + 4\sqrt{5}\ }}{2}.\]

\[t = - 5 - \sqrt{5}:\]

\[x^{2} - 5x = - 5 - \sqrt{5}\]

\[x^{2} - 5x + \left( 5 + \sqrt{5} \right) = 0\]

\[D = 25 - 4 \cdot \left( 5 + \sqrt{5} \right) = 25 -\]

\[- 20 - 4\sqrt{5} = 5 - 4\sqrt{5} < 0\]

\[нет\ корней.\]

\[Ответ:x = \frac{5 \pm \sqrt{5 + 4\sqrt{5}\ }}{2}.\]