ГДЗ по алгебре 8 класс Мерзляк Задание 761

\[\boxed{\mathbf{761\ (761).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1.\frac{9a^{2} - 4}{2a^{2} - 5a + 2} =\]

\[= \frac{(3a - 2)(3a + 2)}{2 \cdot (a - 0,5)(a - 2)}\]

\[2a² - 5a + 2 = 0\]

\[a_{1} + a_{2} = 2,5,\ \ a_{1} = 0,5\]

\[a_{1} \cdot a_{2} = 1,\ \ a_{2} = 2\]

\[2.\ \frac{(3a - 2)(3a + 2)(a - 2)}{2 \cdot (a - 0,5)\left( a - 2)(3a + 2 \right)} =\]

\[= \frac{3a - 2}{2a - 1}\]

\[3.\frac{3a - 2}{2a - 1} - \frac{a - 1}{2a - 1} =\]

\[= \frac{3a - 2 - a + 1}{2a - 1} = \frac{2a - 1}{2a - 1} = 1\]

\[1.\ \frac{b - 1}{2b^{2} + 3b + 1} =\]

\[= \frac{b - 1}{2 \cdot (b + 1)(b + 0,5)}\]

\[2b^{2} + 3b + 1 = 0\]

\[b_{1} + b_{2} = - \frac{3}{2},\ \ b_{1} = - 1\]

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

\[= \frac{b² - 4b}{(2b + 1)(b + 1)(b - 1)}\ \]

\[= \frac{2b + 1}{b^{2}}\ \]

\[1.\frac{c + 2}{c^{2} - c - 6} = \frac{(c + 2)}{(c - 3)(c + 2)} =\]

\[= \frac{1}{c - 3}\]

\[c^{2} - c - 6 = 0\]

\[c_{1} + c_{2} = 1,\ \ c_{1} = 3\]

\[c_{1} \cdot c_{2} = - 6,\ \ c_{2} = - 2\]

\[2.\frac{2c}{c^{2} - 6c + 9} = \frac{2c}{(c - 3)^{2}}\]

\[c^{2} - 6c + 9 = 0\]

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

\[c_{1} \cdot c_{2} = 9,\ \ c_{2} = 3\]

\[3.\frac{1}{c - 3} - \frac{2c}{(c - 3)^{2}} =\]

\[= \frac{c - 3 - 2c}{(c - 3)^{2}} = \frac{- 3 - c}{(c - 3)²}\ \]

\[4.\frac{( - 3 - c)(2c - 6)^{2}}{(c - 3)^{2} \cdot c(c + 3)} =\]

\[= \frac{- (3 + c) \cdot 4 \cdot (c - 3)²}{(c - 3)^{2} \cdot c \cdot (c + 3)} = - \frac{4}{c}\ \]

\[1.\frac{4m - 6}{m^{2} - 3m - 4} =\]

\[= \frac{2 \cdot (2m - 3)}{(m - 4)(m + 1)}\]

\[m^{2} - 3m - 4 = 0\]

\[m_{1} + m_{2} = 3,\ \ m_{1} = 4\]

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

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

\[= \frac{2 \cdot (m - 1,5)(m + 1)}{(m - 4)(m + 1)}\]

\[2m² - m - 3 = 0\]

\[m_{1} + m_{2} = \frac{1}{2},\ \ m_{1} = \frac{3}{2} = 1,5\]

\[m_{1} \cdot m_{2} = - \frac{3}{2},\ \ m_{2} = - 1\]

\[4.\frac{2 \cdot (m - 1,5)(m + 1)(m - 4)}{(m - 4)(m + 1)(2m - 3)} =\]

\[= 4\]