Решебник по алгебре 9 класс Макарычев Задание 973

 

\[\boxed{\text{973\ (973).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} - 2x - 8 = 0\]

\[D = 4 + 32 = 36\]

\[x_{1} = \frac{2 + 6}{2} = 4,\ \ \]

\[x_{2} = \frac{2 - 6}{2} = - 2,\]

\[1)\ x_{1} = 4,\ \ y = 8,\]

\[2)\ x_{2} = - 2,\ \ y_{2} = - 4.\]

\[Ответ:(4;8);\ \ ( - 2;\ - 4).\]

\[Ответ:(5;3).\]

\[x^{2} - 5x + 4 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 + 3}{2} = 4,\ \ \]

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

\[1)\ x_{1} = 4,\ \ y_{1} = 1,\]

\[2)\ x_{2} = 1,\ \ y_{2} = 4.\]

\[Ответ:(4;1);\ \ \ (1;4).\]

\[1)\ y_{1} = 1,\ \ x_{1} = - 3,\]

\[2)\ y_{2} = - 1,\ \ x_{2} = 3.\]

\[Ответ:( - 3;1);\ \ (3;\ - 1).\]

\[4x^{2} - 11x = 0\]

\[x(4x - 11) = 0\]

\[x_{1} = 0,\ \ x_{2} = \frac{11}{4},\]

\[1)\ x_{1} = 0,\ \ y_{1} = 4,\]

\[2)\ x_{2} = \frac{11}{4} = 2\frac{3}{4},\ \ \]

\[y_{2} = \frac{109}{8} = 13\frac{5}{8}.\]

\[Ответ:(0;4);\ \ \left( 2\frac{3}{4};13\frac{5}{8} \right).\]

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

\[D = 1 + 48 = 49,\]

\[y_{1} = \frac{1 + 7}{4} = 2,\ \ \]

\[y_{2} = \frac{1 - 7}{4} = - 1,5;\]

\[1)\ y_{1} = 2,\ \ x_{1} = 1,\]

\[2)\ y_{2} = - 1,5,\ \ x_{2} = \frac{1}{8}.\]

\[Ответ:(1;2);\ \ \left( \frac{1}{8};\ - 1,5 \right).\]