Решебник по алгебре 8 класс Макарычев ФГОС Задание 550

 

\[\boxed{\text{550\ (550).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} - 8x + 9 = 0\]

\[D_{1} = 4^{2} - 9 = 16 - 9 = 7\]

\[x_{1} = 4 + \sqrt{7} \approx 6,65;\]

\[x_{2} = 4 - \sqrt{7} \approx 1,35.\]

\[Ответ:x = 6,65;\ \ x = 1,35.\]

\[\textbf{б)}\ 2y^{2} - 8y + 5 = 0\]

\[D_{1} = 4^{2} - 2 \cdot 5 = 16 - 10 = 6\]

\[y_{1} = \frac{4 + \sqrt{6}}{2} \approx 3,22;\ \ \]

\[y_{2} = \frac{4 - \sqrt{6}}{2} \approx 0,78.\ \]

\[Ответ:y = 0,78;\ \ y = 3,22.\]

\[\boxed{\text{550.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Перенесем все числа влево, приравняем уравнение к нулю.

Избавимся от дробных чисел: умножим на НОК знаменателей все слагаемые.

Решение.

\[\textbf{а)}\ \frac{1}{7}x^{2} = 2x - 7\ \ \ \ \ \ | \cdot 7\]

\[x^{2} - 14x + 49 = 0\]

\[D = 49 - 49 = 0\]

\[x = \frac{7}{1} = 7\]

\[Ответ:x = 7.\]

\[\textbf{б)}\ x^{2} + \frac{6}{5} = 2,6x\ \ \ \ \ \ \ \ \ | \cdot 5\]

\[5x^{2} - 13x + 6 = 0\]

\[D = 169 - 120 = 49\]

\[x_{1} = \frac{13 + 7}{10} = 2;\ \ \ \ \ \]

\[x_{2} = \frac{13 - 7}{10} = \frac{6}{10} = 0,6\]

\[Ответ:x = 2;\ \ x = 0,6.\]

\[\textbf{в)}\ 4x^{2} = 7x + 7,5\ \ \ \ \ \ \ \ | \cdot 2\]

\[8x^{2} - 14x - 15 = 0\]

\[D = 49 + 120 = 169\]

\[x_{1} = \frac{7 + 13}{8} = \frac{20}{8} = 2,5;\ \ \ \ \]

\[\ x_{2} = \frac{7 - 13}{8} = - \frac{6}{8} = - \frac{3}{4} =\]

\[= - 0,75\]

\[Ответ:x = 2,5;\ \ \ x = - 0,75.\]

\[\textbf{г)}\ 6x^{2} - 2 = x\ \]

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

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3};\ \ \ \]

\[\ x_{2} = \frac{1 - 7}{12} = - \frac{6}{12} = - \frac{1}{2} =\]

\[= - 0,5\]

\[Ответ:x = \frac{2}{3};\ \ x = - 0,5.\]