ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 768

\[\boxed{\text{768.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[ax^{2} + bx + c = 0,\ \ \]

\[a + b + c = 0\]

\[D = b^{2} - 4ac = a^{2} + 2ac + c^{2} -\]

\[- 4ac = a^{2} - 2ac + c^{2} =\]

\[= (a - c)^{2}\]

\[x_{1,2} = \frac{- b \pm \sqrt{(a - c)^{2}}}{2a} =\]

\[= \frac{- b \pm |a - c|}{2a} =\]

\[= \frac{a + c \pm (a - c)}{2a}\]

\[x_{1} = \frac{a + c + a - c}{2a} = \frac{2a + 0}{2a} = 1\]

\[x_{2} = \frac{a + c - a + c}{2a} = \frac{2c}{2a} = \frac{c}{a}\]

\[Что\ и\ требовалось\ доказать.\]

\[\textbf{а)}\ 2x^{2} - 41x + 39 = 0\]

\[2 + ( - 41) + 39 = 0 \Longrightarrow x_{1} = 1\]

\[x^{2} - 20,5x + 19,5 = 0\]

\[по\ теореме\ Виета:\]

\[x_{1}x_{2} = 19,5 \Longrightarrow x_{2} = 19,5\]

\[Ответ:x_{1} = 1;\ x_{2} = 19,5.\]

\[\textbf{б)}\ 17x^{2} + 243x - 260 = 0\]

\[17 + 247 + ( - 260) =\]

\[= 0 \Longrightarrow x_{1} = 1\]

\[x^{2} + \frac{243}{17}x - \frac{260}{17} = 0\]

\[по\ теореме\ Виета:\]

\[x_{1}x_{2} = - \frac{260}{17}\]

\[x_{2} = - \frac{260}{17}\]

\[Ответ:x_{1} = 1;\ \ x_{2} = - 15\frac{5}{17}\text{.\ }\]