Решебник по алгебре 9 класс Мерзляк Задание 131

 

\[\boxed{\text{131\ (131).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[11k - 3 \geq 15k - 13\]

\[11k - 15k \geq - 13 + 3\]

\[- 4k \geq - 10\]

\[k \leq 2,5\]

\[Ответ:при\ k \in ( - \infty;2,5\rbrack.\]

\[\boxed{\mathbf{131.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x^{2} - 3x - 4 = 0\]

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = - 1;\ \ \ x_{2} = 4.\]

\[2)\ 2x^{2} + 7x - 4 = 0\]

\[D = 49 + 32 = 81\]

\[x_{1} = \frac{- 7 + 9}{4} = \frac{2}{4} = 0,5;\]

\[x_{2} = \frac{- 7 - 9}{4} = - \frac{16}{4} = - 4.\]

\[3)\ 6x^{2} - 8x + 3 = 0\]

\[D_{1} = 16 - 18 = - 2 < 0\]

\[нет\ корней.\]

\[4)\ 3x^{2} + x - 2x = - 7x^{2}\]

\[10x^{2} - x = 0\]

\[10x\left( x - \frac{1}{10} \right) = 0\]

\[x = 0;\ \ x = 0,1.\]

\[5)\ (x + 1)^{2} = 2x^{2} + 6x - 31\]

\[x^{2} + 2x + 1 = 2x^{2} + 6x - 31\]

\[x^{2} + 4x - 30 = 0\]

\[D_{1} = 4 + 30 = 34\]

\[x = - 2 \pm \sqrt{34}.\]

\[6)\ (4x - 1)(x - 2) = 3\]

\[4x^{2} - x - 8x + 2 - 3 = 0\]

\[4x^{2} - 9x - 1 = 0\]

\[D = 81 + 16 = 97\]

\[x = \frac{9 \pm \sqrt{97}}{8}.\]