Решебник по алгебре 9 класс Макарычев Задание 31

 

\[\boxed{\text{31\ (31).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} + 7x + 12 = 0\]

\[D = 7^{2} - 4 \cdot 12 = 49 - 48 = 1\]

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

\[= \frac{- 7 - 1}{2} = - 4.\]

\[Ответ:x = - 3;x = - 4.\]

\[\textbf{б)}\ x^{2} - 2x - 35 = 0\]

\[D_{1} = 1 + 35 = 36\]

\[x_{1} = 1 + 6 = 7;\ \ \ x_{2} = 1 - 6 =\]

\[= - 5.\]

\[Ответ:x = 7;x = - 5.\]

\[\textbf{в)}\ 2x^{2} - 5x - 3 = 0\]

\[D = ( - 5)^{2} + 4 \cdot 2 \cdot 3 =\]

\[= 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{4} = 3;\ \ \ \ x_{2} = \frac{5 - 7}{4} =\]

\[= - \frac{2}{4} = - 0,5.\]

\[Ответ:x = 3;x = - 0,5.\]

\[\textbf{г)}\ 3x^{2} - 8x + 5 = 0\]

\[D_{1} = ( - 4)^{2} - 3 \cdot 5 = 16 - 15 = 1\]

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

\[= \frac{4 - 1}{3} = 1.\]

\[Ответ:x = 1;\ \ x = 1\frac{2}{3}.\]

\[\boxed{\text{31.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ c > 1:\]

\[c > \sqrt{c}.\]

\[\textbf{б)}\ 0 < c < 1:\]

\[c < \sqrt{c}.\]

\[Существует:\]

\[c = 0;\ \ c = 1.\]