ГДЗ по алгебре 8 класс Макарычев Задание 672

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\[\textbf{а)}\ 2x^{2} + bx - 10 = 0,\ \ x_{1} = 5\]

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

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

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

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

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

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

\[b = - 17.\]

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

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

\[2b = 28\]

\[b = 14\]

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

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

\[\ b = 14.\]

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

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

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

\[3b = 1\]

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

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

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

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

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