Решебник по алгебре 8 класс Мерзляк ФГОС Задание 756

 

\[\boxed{\mathbf{756\ (756).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[= \frac{(x - 5)(x - 1)}{(x - 5)} = x - 1\]

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

\[x_{1} + x_{2} = 6,\ \ x_{1} = 5\]

\[x_{1} \cdot x_{2} = 5,\ \ x_{2} = 1\]

\[2)\ \frac{2x + 12}{x^{2} + 3x - 18} =\]

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

\[x^{2} + 3x - 18 = (x + 6)(x - 3)\]

\[x_{1} + x_{2} = - 3,\ \ x_{1} = - 6\]

\[x_{1} \cdot x_{2} = - 18,\ \ x_{2} = 3\]

\[3)\frac{x^{2} + 9x + 14}{x^{2} + 7x} =\]

\[= \frac{(x + 7)(x + 2)}{x(x + 7)} = \frac{x + 2}{x}\]

\[x^{2} + 9x + 14 = (x + 7)(x + 2)\]

\[x_{1} + x_{2} = - 9,\ \ x_{1} = - 7\]

\[x_{1} \cdot x_{2} = 14,\ \ x_{2} = - 2\]

\[\ \]

\[\boxed{\mathbf{7}\mathbf{5}\mathbf{6}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[2x^{2} - 7x - 13 = 0\]

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

\[= x_{1} \cdot x_{2} - 4 \cdot \left( x_{1} + x_{2} \right) =\]

\[= - \frac{13}{2} - 4 \cdot \left( \frac{7}{2} \right) =\]

\[= - \frac{13}{2} - \frac{28}{2} = - \frac{41}{2} = - 20,5.\]