ГДЗ по алгебре 9 класс Макарычев Задание 78

\[\boxed{\text{78\ (78).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 2x^{2} + 12x - 14 =\]

\[= 2 \cdot (x^{2} + 6x - 7)\]

\[x^{2} + 6x - 7 = 0\]

\[D_{1} = 3^{2} + 7 = 16\]

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

\[= - 3 - 4 = - 7.\]

\[2x^{2} + 12x - 14 =\]

\[= 2 \cdot (x - 1)(x + 7).\]

\[\textbf{б)} - m^{2} + 5m - 6 =\]

\[= - (m^{2} - 5m + 6)\]

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

\[D = 5^{2} - 4 \cdot 6 = 25 - 24 = 1\]

\[m_{1} = \frac{5 + 1}{2} = 3;\ m_{2} = \frac{5 - 1}{2} = 2.\]

\[- m^{2} + 5m - 6 =\]

\[= - (m - 3)(m - 2) =\]

\[= (3 - m)(m - 2).\]

\[\textbf{в)}\ 3x^{2} + 5x - 2 = 0\]

\[D = 5^{2} + 4 \cdot 3 \cdot 2 = 25 + 24 = 49\]

\[x_{1} = \frac{- 5 - 7}{6} = - 2;\ \ \ \ \ \ x_{2} =\]

\[= \frac{- 5 + 7}{6} = \frac{2}{6} = \frac{1}{3}.\]

\[3x^{2} + 5x - 2 =\]

\[= 3 \cdot (x + 2)\left( x - \frac{1}{3} \right) =\]

\[= (x + 2)(3x - 1)\]

\[\textbf{г)}\ 6x^{2} - 13x + 6 = 0\]

\[D = 13^{2} - 4 \cdot 6 \cdot 6 =\]

\[= 169 - 144 = 25\]

\[x_{1} = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2};\ \ \ \ \ \ x_{2} =\]

\[= \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3}.\]

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

\[= 6 \cdot \left( x - \frac{3}{2} \right)\left( x - \frac{2}{3} \right) =\]

\[= (2x - 3)(3x - 2)\ \]