Решебник по алгебре 8 класс Мерзляк Задание 757

\[\boxed{\mathbf{757\ (757).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{4a^{2} - 9}{2a^{2} - 9a - 18} =\]

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

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

\[2a^{2} - 9a - 18 =\]

\[= 2 \cdot (a - 6)(a + 1,5) =\]

\[= (a - 6)(2a + 3)\]

\[a_{1} + a_{2} = 4,5,\ \ a_{1} = 6\]

\[a_{1} \cdot a_{2} = - 9,\ \ a_{2} = - 1,5\]

\[2)\ \frac{2b^{2} - 7b + 3}{4b^{2} - 4b + 1} =\]

\[= \frac{2 \cdot \left( b - \frac{1}{2} \right)(b - 3)}{4 \cdot \left( b - \frac{1}{2} \right)^{2}} =\]

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

\[3)\ \frac{c^{2} - 5c - 6}{c^{2} - 8c + 12} =\]

\[= \frac{(c - 6)(c + 1)}{(c - 6)(c - 2)} = \frac{c + 1}{c - 2}\]

\[4)\ \frac{m^{3} - 1}{m^{2} + 9m - 10} =\]

\[= \frac{(m - 1)\left( m^{2} + m + 1 \right)}{(m - 1)(m + 10)} =\]

\[= \frac{m^{2} + m + 1}{m + 10}\]

\[m^{2} + 9m - 10 =\]

\[= (m + 10)(m - 1)\]

\[m_{1} + m_{2} = - 9,\ \ m_{1} = - 10\]

\[m_{1} \cdot m_{2} = - 10,\ \ m_{2} = 1\]

\[5)\ \frac{x^{2} - 16}{32 - 4x - x^{2}} =\]

\[= \frac{(x - 4)(x + 4)}{- (x - 4)(x + 8)} = \frac{x + 4}{- x - 8}\]

\[- x^{2} - 4x + 32 = (x + 8)(x - 4)\]

\[x_{1} + x_{2} = - 4,\ \ x_{1} = - 8\]

\[x_{1} \cdot x_{2} = - 32,\ \ x_{2} = 4\]

\[6)\ \frac{4n^{2} - 9n + 2}{2 + 9n - 5n^{2}} =\]

\[= \frac{(4n - 1)(n - 2)}{- (n - 2)(5n + 5)} = \frac{4n - 1}{- 5n - 1}\]

\[4n^{2} - 9n + 2 =\]

\[= 4 \cdot \left( n - \frac{1}{4} \right)(n - 2) =\]

\[= (4n - 1)(n - 2)\]

\[n_{1} + n_{2} = \frac{9}{4},\ \ n_{1} = \frac{1}{4}\]

\[n_{1} \cdot n_{2} = \frac{2}{4},\ \ n_{2} = \frac{8}{4} = 2\]

\[- 5n^{2} + 9n + 2 =\]

\[= - 5 \cdot (n - 2)\left( n + \frac{1}{5} \right) =\]

\[= - (n - 2)(5n + 5)\]

\[n_{1} + n_{2} = \frac{9}{5},\ \ n_{1} = \frac{10}{5} = 2\]

\[n_{1} \cdot n_{2} = - \frac{2}{5},\ \ n_{2} = - \frac{1}{5}\]