ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 765

\[\boxed{\text{765.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ 2x^{2} + bx - 10 = 0,\]

\[\text{\ \ }x_{1} = 5\]

\[x_{2} + \frac{b}{2}x - 5 = 0\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = - \frac{b}{2} \\ x_{1}x_{2} = - 5\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} 5 + x_{2} = - \frac{b}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 5x_{2} = - 5 \Longrightarrow x_{2} = - 1 \\ \end{matrix} \right.\ \]

\[- \frac{b}{2} = 5 + ( - 1) = 4\ \ \ \ \]

\[\ | \cdot ( - 2) \Longrightarrow b = - 8\]

\[Ответ:x_{1} = 5,\ x_{2} = - 1,\ b = - 8\]

\[\textbf{б)}\ 3x^{2} + bx + 24 = 0,\ \ \]

\[x_{1} = 3\]

\[x_{2} + \frac{b}{3}x + 8 = 0\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = - \frac{b}{3} \\ x_{1}x_{2} = 8\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} 3 + x_{2} = - \frac{b}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 3x^{2} = 8 \Longrightarrow x_{2} = \frac{8}{3} \\ \end{matrix} \right.\ \ \]

\[- \frac{b}{3} = 3 + \frac{8}{3} = \frac{9 + 8}{3} =\]

\[= \frac{17}{3} \Longrightarrow b = - 17\]

\[Ответ:x_{1} = 3,\ x_{2} = 2\frac{2}{3},\]

\[\ b = - 17.\]

\[\textbf{в)}\ (b - 1)x^{2} - (b + 1)x = 72,\]

\[\text{\ \ }x_{1} = 3\]

\[x^{2} - \frac{b + 1}{b - 1}x - \frac{72}{b - 1} = 0\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = \frac{b + 1}{b - 1} \\ x_{1}x_{2} = - \frac{72}{b - 1} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3 + x_{2} = \frac{b + 1}{b - 1}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 3x^{2} = - \frac{72}{b - 1}\ \ \ |\ :3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 3 + x_{2} = \frac{b + 1}{b - 1} \\ x_{2} = - \frac{24}{b - 1}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

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

\[+ \left( - \frac{24}{b - 1} \right)\ \ \ \ \ | \cdot (b - 1)\]

\[b + 1 = 3b - 3 - 24\]

\[2b = 28\]

\[b = 14\]

\[x_{2} = - \frac{24}{14 - 1} = - \frac{24}{13}\]

\[Ответ:x_{1} = 3,\ x_{2} = - 1\frac{1}{13},\ \]

\[b = 14.\]

\[\textbf{г)}\ (b - 5)x^{2} - (b - 2)x + b =\]

\[= 0,\ \ x_{1} = \frac{1}{2}\]

\[x^{2} - \frac{b - 2}{b - 5}x + \frac{b}{b - 5} = 0\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = \frac{b - 2}{b - 5} \\ x_{1}x_{2} = \frac{b}{b - 5}\text{\ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \frac{1}{2} + x_{2} = \frac{b - 2}{b - 5} \\ \frac{1}{2}x_{2} = \frac{b}{b - 5}\ \ \ | \cdot 2 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} \frac{1}{2} + x_{2} = \frac{b - 2}{b - 5} \\ x_{2} = \frac{2b}{b - 5}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\frac{1}{2} + \frac{2b}{b - 5} = \frac{b - 2}{b - 5}\ \ \ \ | \cdot 2(b - 5)\]

\[b - 5 + 4b = 2b - 4\]

\[3b = 1\]

\[b = \frac{1}{3}\]

\[x_{2} = \frac{2b}{b - 5} = \frac{2 \cdot \frac{1}{3}}{\frac{1}{3} - 5} = \frac{\frac{2}{3}}{\frac{1 - 15}{3}} =\]

\[= - \frac{2}{14} = - \frac{1}{7}\]

\[Ответ:x_{1} = 0,5;\ \ \ x_{2} = - \frac{1}{7};\ \ \ \]

\[b = \frac{1}{3}\text{.\ }\]