Решебник по алгебре 8 класс Макарычев ФГОС Задание 696

 

\[\boxed{\text{696\ (696).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[16x^{2} + 40x - 52 = 0\ \ \ \ \ \ |\ :4\]

\[4x^{2} + 10x - 13 = 0\]

\[D = 100 + 208 = 308\]

\[x_{1,2} = \frac{- 10 \pm \sqrt{308}}{8} = \frac{- 5 \pm \sqrt{77}}{4}\]

\[Ответ:x = \frac{- 5 \pm \sqrt{77}}{4}.\]

\[y \neq 0\]

\[y^{2} \neq 9,\ \ y \neq \pm 3\]

\[2y^{2} - 2y + 6 + 3y - 9 = 0\]

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

\[D = 1 + 24 = 25\]

\[y_{1,2} = \frac{- 1 \pm 5}{4} = 1;\ - \frac{3}{2}\]

\[Ответ:y = \left\{ - 1,5;1 \right\}.\]

\[\textbf{в)}\frac{2y - 1}{14y^{2} + 7y} + \frac{8}{12y^{2} - 3} =\]

\[= \frac{2y + 1}{6y^{2} - 3y}\]

\[y \neq 0\]

\[4y^{2} \neq 1,\ \ y^{2} \neq \frac{1}{4},\]

\[\ \ y \neq \pm \frac{1}{2}\]

\[3 \cdot (2y - 1)^{2} + 56y =\]

\[= 7 \cdot (2y + 1)^{2}\]

\[12y^{2} - 12y + 3 + 56y =\]

\[= 28y^{2} + 28y + 7\]

\[16y^{2} - 16y + 4 = 0\ \ \ \ \ \ |\ :4\]

\[4y^{2} - 4y + 1 = 0\]

\[D = 16 - 16 = 0\]

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

\[2y = 1\]

\[y = \frac{1}{2} - не\ подходит\ по\ ОДЗ\]

\[Ответ:корней\ нет.\]

\[\textbf{г)}\frac{3}{x^{2} - 9} - \frac{1}{9 - 6x + x^{2}} =\]

\[= \frac{3}{2x^{2} + 6x}\]

\[x \neq 0\]

\[x^{2} \neq 9,\ \ x \neq \pm 3\]

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

\[= 3 \cdot (x - 3)^{2}\]

\[6x^{2} - 18x - 2x^{2} - 6x =\]

\[= 3x^{2} - 18x + 27\]

\[x^{2} - 6x - 27 = 0\]

\[D = 36 + 108 = 144\]

\[x_{1,2} = \frac{6 \pm 12}{2} = 9;\ - 3 -\]

\[не\ подходит\ по\ ОДЗ\]

\[Ответ:x = 9.\]

\[x^{3} \neq 64,\ \ x \neq 4\]

\[9x + 12 - x + 4 = x^{2} + 4x + 16\]

\[x^{2} - 4x = 0\]

\[x(x - 4) = 0\]

\[x = 0,\]

\[\ \ x = 4 - не\ подходит\ по\ ОДЗ\]

\[Ответ:x = 0.\]

\[y^{3} \neq - \frac{1}{8},\ \ y \neq - \frac{1}{2}\]

\[3 - 4y^{2} + 2y - 1 =\]

\[= (y + 3)(2y + 1)\]

\[4y^{2} + 2y + 2 =\]

\[= 2y^{2} + y + 6y + 3\]

\[6y^{2} + 5y + 1 = 0\]

\[D = 25 - 24 = 1\]

\[y_{1,2} = \frac{- 5 \pm 1}{12} = - \frac{1}{3};\ - \frac{1}{2} -\]

\[не\ подходит\ по\ ОДЗ\]

\[Ответ:y = - \frac{1}{3}.\]

\[x \neq 2,\ \ x \neq \pm 1\]

\[32 + x + 1 = (x - 2)(x - 1)\]

\[x + 33 = x^{2} - x - 2x + 2\]

\[x^{2} - 4x - 31 = 0\]

\[D = 16 + 124 = 140\]

\[x_{1,2} = \frac{4 \pm \sqrt{140}}{2} = \frac{4 \pm 2\sqrt{35}}{2} =\]

\[= 2 \pm \sqrt{35}\]

\[Ответ:x = 2 \pm \sqrt{35}.\]

\[x \neq 4\]

\[2 \cdot \left( x^{2} + 3 \right) + 3 \cdot (x - 4) + 6 = 0\]

\[2x^{2} + 6 + 3x - 12 + 6 = 0\]

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

\[x(2x + 3) = 0\]

\[x = 0,\ \ 2x + 3 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{3}{2}\]

\[Ответ:x = \left\{ - 1,5;0 \right\}\text{.\ }\]

\[\boxed{\text{696.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\left\{ \begin{matrix} x² + y² = 5\ \ \ \ \\ 6x + 5y = - 4 \\ \end{matrix} \right.\ \]

\[\textbf{а)}\ ( - 2;1):\ \ \]

\[\left\{ \begin{matrix} ( - 2)^{2} + 1² = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6 \cdot ( - 2) + 5 \cdot 1 = - 7 \neq - 4 \\ \end{matrix} \right.\ \]

\[\Longrightarrow ( - 2;1)\ не\ является\]

\[\ решением\ системы;\]

\[\textbf{б)}\ (1;\ - 2):\ \ \]

\[\left\{ \begin{matrix} 1^{2} + ( - 2)^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \\ 6 \cdot 1 + 5 \cdot ( - 2) = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 1 + 4 = 5\ \ \ \ \ \\ 6 - 10 = - 4 \\ \end{matrix} \right.\ \]

\[\Longrightarrow (1;\ - 2)\ является\ \]

\[решением\ системы.\]