ГДЗ по алгебре 8 класс Мерзляк Задание 795

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\[\left\{ \begin{matrix} x^{2} - 6x = t\ \ \ \ \ \\ t^{2} + t - 56 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = - 1,\ \ t_{1} \cdot t_{2} = - 56,\ \]

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

\[Ответ:x = 4;x = 2;x = 7;\ \]

\[x = - 1.\]

\[2)\ \left( x^{2} + 8x + 3 \right)(x² + 8x + 5) =\]

\[= 63\]

\[\left\{ \begin{matrix} x^{2} + 8x + 3 = t \\ t^{2} + 2t - 63 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = - 2,\ \ t_{1} \cdot t_{2} = - 63,\ \ \]

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

\[Ответ:\ x = - 6;\ x = - 2;\ \]

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

\[3)\ \frac{x^{4}}{(x - 2)²} - \frac{4x^{2}}{x - 2} - 5 = 0\]

\[\left\{ \begin{matrix} \frac{x^{2}}{x - 2} = t \\ t^{2} - 4t - 5 = 0 \\ \end{matrix} \right.\ \]

\[t_{1} + t_{2} = 4,\ \ t_{1} \cdot t_{2} = - 5,\ \ \]

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

\[\left\{ \begin{matrix} \frac{x^{2}}{x - 2} = 5 \\ \frac{x^{2}}{x - 2} = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[Ответ:\ x = - 2;x = 1.\]

\[4)\ \frac{x + 4}{x - 3} - \frac{x - 3}{x + 4} = \frac{3}{2}\ \]

\[\left\{ \begin{matrix} \frac{x + 4}{x - 3} = t \\ t - \frac{1}{t} - \frac{3}{2} = 0 \\ \end{matrix} \right.\ \]

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

\[D = 9 + 16 = 25,\ \]

\[\ t = \frac{3 - 5}{4} = - 0,5,\]

\[t = \frac{3 + 5}{4} = 2\]

\[Ответ:\ x = - \frac{5}{3};x = 10.\]