Решебник по алгебре 8 класс Мерзляк ФГОС Задание 793

 

\[\boxed{\mathbf{793\ (793).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( x^{2} - 2 \right)^{2} - 8 \cdot \left( x^{2} - 2 \right) + 7 =\]

\[= 0\]

\[\left\{ \begin{matrix} x^{2} - 2 = t\ \ \ \ \ \ \ \ \\ t^{2} - 8t + 7 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 8,\ \ t_{1} \cdot t_{2} = 7,\ \ \]

\[t_{1} = 7,\ \ t_{2} = 1\]

\[Ответ:\ x = \pm 3;\ x = \pm \sqrt{3}.\]

\[\left\{ \begin{matrix} x^{2} + 5x = t\ \ \ \ \ \ \ \ \\ t² - 2t - 24 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 2,\ \ t_{1} \cdot t_{2} = - 24,\ \ \]

\[t_{1} = 6,\ \ t_{2} = - 4\]

\[Ответ:\ x = - 6;\ x = - 4;x = 1;\ \]

\[x = - 1.\]

\[3)\ \left( x^{2} - 3x + 1 \right)\left( x^{2} - 3x + 3 \right) =\]

\[= 3\]

\[\left\{ \begin{matrix} x² - 3x + 1 = t \\ t(t + 2) - 3 = 0 \\ \end{matrix} \right.\ \]

\[t^{2} + 2t - 3 = 0\]

\[t_{1} + t_{2} = - 2,\ \ t_{1} \cdot t_{2} = - 3,\ \ \]

\[t_{1} = - 3,\ \ t_{2} = 1\]

\[\ x_{1} + x_{2} = 3,\ \ x_{1}x_{2} = 4\ \]

\[нет\ корней.\]

\[x(x - 3) = 0\]

\[x = 0,\ \ x = 3\]

\[Ответ:x = 0;x = 3.\]

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

\[= - 5\]

\[\left\{ \begin{matrix} x^{2} + 2x + 2 = t \\ t(t - 6) + 5 = 0 \\ \end{matrix} \right.\ \]

\[t^{2} - 6t + 5 = 0\]

\[t_{1} + t_{2} = 6,\ \ t_{1} \cdot t_{2} = 5,\ \]

\[\ t_{1} = 5,\ \ t_{2} = 1\]

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

\[x_{1} = - 3,\ \ x_{2} = 1\]

\[Ответ:\ x = - 1;x = 1;\ x = - 3.\]

\[\boxed{\mathbf{7}\mathbf{9}\mathbf{3}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{x^{2} - 6x + 5}{x - 5} =\]

\[= \frac{(x - 5)(x - 1)}{(x - 5)} = x - 1\]

\[x^{2} - 6x + 5 = (x - 5)(x - 1)\]

\[x_{1} + x_{2} = 6,\ \ x_{1} = 5\]

\[x_{1} \cdot x_{2} = 5,\ \ x_{2} = 1\]

\[2)\ \frac{2x + 12}{x^{2} + 3x - 18} =\]

\[= \frac{2 \cdot (x + 6)}{(x + 6)(x - 3)} = \frac{2}{x - 3}\]

\[x^{2} + 3x - 18 = (x + 6)(x - 3)\]

\[x_{1} + x_{2} = - 3,\ \ x_{1} = - 6\]

\[x_{1} \cdot x_{2} = - 18,\ \ x_{2} = 3\]

\[3)\frac{x^{2} + 9x + 14}{x^{2} + 7x} =\]

\[= \frac{(x + 7)(x + 2)}{x(x + 7)} = \frac{x + 2}{x}\]

\[x^{2} + 9x + 14 = (x + 7)(x + 2)\]

\[x_{1} + x_{2} = - 9,\ \ x_{1} = - 7\]

\[x_{1} \cdot x_{2} = 14,\ \ x_{2} = - 2\]

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