Решебник по алгебре 9 класс Мерзляк Задание 249

 

\[\boxed{\text{249\ (249).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ x^{2} - x - 12 =\]

\[= (x - 4) \cdot (x + 3)\]

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

\[x_{1}x_{2} = - 12,\ \ x_{2} = - 3.\]

\[2) - x^{2} + 2x + 35 =\]

\[= - \left( x^{2} - 2x - 35 \right) =\]

\[= - (x - 7) \cdot (x + 5)\]

\[x_{1} + x_{2} = 2,\ \ x_{1} = 7\]

\[x_{1}x_{2} = - 35,\ \ x_{2} = - 5.\]

\[3)\ 6x^{2} + 11x - 2\]

\[D = 121 + 48 = 169\]

\[x = \frac{- 11 \pm 13}{12}\]

\[x = - 2\]

\[x = \frac{1}{6}\]

\[6x^{2} + 11x - 2 =\]

\[= 6 \cdot (x + 2) \cdot \left( x - \frac{1}{6} \right) =\]

\[= (x + 2)(6x - 1).\]

\[4)\ \frac{2}{3}x^{2} + 3x - 6\]

\[D = 9 + 16 = 25\]

\[x = \frac{- 3 \pm 5}{\frac{4}{3}}\]

\[x = - 6\]

\[x = 1,5\]

\[\frac{2}{3}x^{2} + 3x - 6 =\]

\[= \frac{2}{3} \cdot (x + 6) \cdot (x - 1,5).\]

\[\boxed{\mathbf{249.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1) - 8x \leq - 24\]

\[x \geq 3.\]

\[2)\ \frac{3}{7}x > 21\]

\[x > 21 \cdot \frac{7}{3}\]

\[x > 49.\]

\[3) - 14x \leq 0\]

\[x \geq 0.\]

\[4) - 3x < \frac{9}{11}\]

\[x > \frac{9}{11}\ :( - 3)\]

\[x > - \frac{3}{11}.\]

\[5)\ 12 - 5x \geq - 23\]

\[- 5x \geq - 35\]

\[x \leq 7.\]

\[6)\ \frac{3 - 2x}{5} < 1\]

\[3 - 2x < 5\]

\[- 2x < 2\]

\[x > - 1.\]