Решебник по алгебре 8 класс Мерзляк ФГОС Задание 795

 

\[\boxed{\mathbf{795\ (795).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

\[Ответ:x = 4;x = 2;x = 7;\ \]

\[x = - 1.\]

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

\[= 63\]

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

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

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

\[Ответ:\ x = - 6;\ x = - 2;\ \]

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

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

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

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

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

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

\[Ответ:\ x = - 2;x = 1.\]

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

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

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

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

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

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

\[Ответ:\ x = - \frac{5}{3};x = 10.\]

\[\boxed{\mathbf{7}\mathbf{9}\mathbf{5}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{4x^{2} + x - 3}{x^{2} - 1} =\]

\[= \frac{(x + 1)(4x - 3)}{(x - 1)(x + 1)} = \frac{4x - 3}{x - 1}\]

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

\[= 4 \cdot (x + 1)\left( x - \frac{3}{4} \right) =\]

\[= (x + 1)(4x - 3)\]

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

\[x_{1} = - \frac{4}{4} = - 1\]

\[x_{1} \cdot x_{2} = - \frac{3}{4},\ \ x_{2} = \frac{3}{4}\]

\[2)\ \frac{2y^{2} + 3y - 5}{y^{2} - 2y + 1} =\]

\[= \frac{(2y + 5)(y - 1)}{(y - 1)^{2}} = \frac{2y + 5}{y - 1}\]

\[2y^{2} + 3y - 5 =\]

\[= 2 \cdot \left( y + \frac{5}{2} \right)(y - 1) =\]

\[= (2y + 5)(y - 1)\]

\[y_{1} + y_{2} = - \frac{3}{2},\ \ y_{1} = - \frac{5}{2}\]

\[y_{1} \cdot y_{2} = - \frac{5}{2},\ \ y_{2} = \frac{2}{2} = 1\]

\[y^{2} - 2y + 1 = 0\]

\[y_{1} + y_{2} = 2,\ \ y_{1} = 1\]

\[y_{1} \cdot y_{2} = 1,\ \ y_{2} = 1\]

\[3)\ \frac{a^{2} + 5a + 4}{a^{2} - a - 20} =\]

\[= \frac{(a + 4)(a + 1)}{(a + 4)(a - 5)} = \frac{a + 1}{a - 5}\]

\[4)\ \frac{3 + 20b - 7b^{2}}{7b^{2} - 6b - 1} =\]

\[= \frac{(3 - b)(7b + 1)}{(7b + 1)(b - 1)} = \frac{3 - b}{b - 1}\]

\[- 7b^{2} + 20b + 3 =\]

\[= - 7 \cdot (b - 3)\left( b + \frac{1}{7} \right) =\]

\[= (3 - b)(7b + 1)\]

\[b_{1} + b_{2} = \frac{20}{7},\ \ b_{1} = \frac{21}{7} = 3\]

\[b_{1} \cdot b_{2} = - \frac{3}{7},\ \ b_{2} = - \frac{1}{7}\]

\[7b^{2} - 6b - 1 =\]

\[= 7 \cdot \left( b + \frac{1}{7} \right)(b - 1) =\]

\[= (7b + 1)(b - 1)\]

\[b_{1} + b_{2} = \frac{6}{7},\ \ b_{1} = - \frac{1}{7}\]

\[b_{1} \cdot b_{2} = - \frac{1}{7},\ \ b_{2} = \frac{7}{7} = 1\]