Решебник по алгебре 8 класс Макарычев ФГОС Задание 672

 

\[\boxed{\text{672\ (672).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

\[b = - 17.\]

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

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

\[2b = 28\]

\[b = 14\]

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

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

\[\ b = 14.\]

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

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

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

\[3b = 1\]

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

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

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

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

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

\[\boxed{\text{672.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ x - 2y = 8\]

\[\Longrightarrow (2;\ - 3);(8;0);(0;\ - 4).\]

\[\textbf{б)}\ x + 0 \cdot y = 10\]

\[x = 10;\ \ y - любое:\]

\[\Longrightarrow (10;100);(10;200);\ \ \]

\[(10;201).\]

\[\textbf{в)}\ x - xy = 12\]

\[\Longrightarrow (12;0);(4;\ - 2);(6;\ - 1).\]

\[\textbf{г)}\ (x + y)(y - 2) = 0\]

\[\Longrightarrow (1;\ - 1);(2;\ - 2);(3;\ - 3).\]