Решебник по алгебре 8 класс Мерзляк ФГОС Задание 688

 

\[\boxed{\mathbf{688\ (688).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 2x² + 4x - b = 0\]

\[D = 16 - 4 \cdot 2 \cdot ( - b) = 0\]

\[16 + 8b = 0\]

\[b = - \frac{16}{8},\ \ b = - 2\]

\[Ответ:при\ b = - 2.\]

\[2)\ 3x² - bx + 12 = 0\]

\[D = b^{2} - 4 \cdot 3 \cdot 12 = 0\]

\[b^{2} = 144\]

\[b = 12;\ \ b = - 12\]

\[Ответ:\ при\ b = 12;\ \ b = - 12.\]

\[\boxed{\mathbf{6}\mathbf{8}\mathbf{8}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

\[3)\ \frac{4}{c^{2} - 4c} - \frac{c + 4}{c^{2} - 16} =\]

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

\[= \frac{4c + 16 - c^{2} - 4c}{c(c - 4)(c + 4)} =\]

\[= \frac{- c^{2} + 16}{c(c - 4)(c + 4)} =\]

\[= \frac{(4 - c)(4 + c)}{c(c - 4)(c + 4)} = - \frac{1}{c}\]

\[4)\ \frac{56a^{5}}{b^{4}} \cdot \frac{b^{2}}{14b^{5}} = \frac{56a^{5} \cdot b^{2}}{b^{4} \cdot 14b^{5}} =\]

\[= \frac{4a^{5}}{b^{7}}\]

\[5)\ \frac{72a^{3}b}{c}\ :\left( 27a^{2}b \right) =\]

\[= \frac{72a^{3}b}{c \cdot 27a^{2}b} = \frac{8a}{3c}\]

\[6)\ \frac{4a^{2} - 1}{a^{2} - 9}\ :\frac{10a + 5}{a + 3} =\]

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

\[= \frac{2a - 1}{5a - 15}\]