Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Макарычев Задание 705
Задание 705
\[\boxed{\text{705\ (705).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ b_{1} = \frac{243}{256},\ \ q = \frac{2}{3},\ \ \]
\[n = 8;\]
\[b_{8} = b_{1} \cdot q^{7} = \frac{243}{256} \cdot \left( \frac{2}{3} \right)^{7} =\]
\[= \frac{3^{5}}{2^{8}} \cdot \frac{2^{7}}{3^{7}} = \frac{1}{18}\text{.\ }\]
\[\textbf{б)}\ b_{1} = \sqrt{\frac{2}{3}},\ \ q = - \sqrt{6},\]
\[n = 5;\]
\[b_{5} = b_{1} \cdot q^{4} = \sqrt{\frac{2}{3}} \cdot \left( - \sqrt{6} \right)^{4} =\]
\[= 36 \cdot \sqrt{\frac{2}{3}} = 12\sqrt{6}.\]
\[\boxed{\text{705.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{21a^{3} - 6a^{2}b}{12ab - 42a^{2}} =\]
\[= \frac{3a²(7a - 2b)}{6a(2b - 7a)} = - \frac{a}{2}\]
\[\textbf{б)}\ \frac{6m^{3} + 3mn^{2}}{2m^{3}n + mn^{3}} =\]
\[= \frac{3m(2m^{2} + n^{2})}{mn(2m^{2} + n^{2})} = \frac{3}{n}\]
\[\textbf{в)}\ \frac{x^{2} - 2mx + 3x - 6m}{x^{2} + 2mx + 3x + 6m} =\]
\[= \frac{x(x + 3) - 2m(x + 3)}{x(x + 3) + 2m(x + 3)} =\]
\[= \frac{(x + 3)(x - 2m)}{(x + 3)(x + 2m)} =\]
\[= \frac{x - 2m}{x + 2m}\]
\[\textbf{г)}\ \frac{8ab + 2a - 20b - 5}{4ab - 8b^{2} + a - 2b} =\]
\[= \frac{2a(4b + 1) - 5 \cdot (4b + 1)}{a(4b + 1) - 2b(4b + 1)} =\]
\[= \frac{(4b + 1)(2a - 5)}{(4b + 1)(a - 2b)} =\]
\[= \frac{2a - 5}{a - 2b}\]
\[\textbf{д)}\ \frac{16a^{2} - 8ab + b^{2}}{16a^{2} - b^{2}} =\]
\[= \frac{(4a - b)²}{(4a - b)(4a + b)} = \frac{4a - b}{4a + b}\]
\[\textbf{е)}\ \frac{9x^{2} - 25y^{2}}{9x^{2} + 30xy + 25y^{2}} =\]
\[= \frac{(3x - 5y)(3x + 5y)}{(3x + 5y)^{2}} =\]
\[= \frac{3x - 5y}{3x + 5y}\]
\[\textbf{ж)}\ \frac{a^{2} - 3a}{a^{2} + 3a - 18} =\]
\[= \frac{a(a - 3)}{(a - 3)(a + 6)} = \frac{a}{a + 6}\]
\[a^{2} + 3a - 18 = (a - 3)(a + 6)\]
\[D = 9 + 72 = 81\]
\[a_{1} = \frac{- 3 + 9}{2} = 3,\ \ \]
\[a_{2} = \frac{- 3 - 9}{2} = - 6.\]
\[\textbf{з)}\ \frac{4x^{2} - 8x + 3}{4x^{2} - 1} =\]
\[= \frac{4 \cdot \left( x - \frac{3}{2} \right)\left( x - \frac{1}{2} \right)}{(2x - 1)(2x + 1)} =\]
\[= \frac{(2x - 3)(2x - 1)}{(2x - 1)(2x + 1)} = \frac{2x - 3}{2x + 1}\]
\[4x^{2} - 8x + 3 =\]
\[= 4 \cdot \left( x - \frac{3}{2} \right)\left( x - \frac{1}{2} \right)\]
\[D = 64 - 48 = 16\]
\[x_{1} = \frac{8 + 4}{8} = \frac{12}{8} = \frac{3}{2},\ \ \]
\[x_{2} = \frac{8 - 4}{8} = \frac{4}{8} = \frac{1}{2}.\]
\[\textbf{и)}\ \frac{m^{2} + 4m - 5}{m^{2} + 7m + 10} =\]
\[= \frac{(m - 1)(m + 5)}{(m + 2)(m + 5)} = \frac{m - 1}{m + 2}\]
\[m^{2} + 4m - 5 = (m - 1)(m + 5)\]
\[D = 16 + 20 = 36\]
\[m_{1} = \frac{- 4 + 6}{2} = 1,\]
\[\text{\ \ }m_{2} = \frac{- 4 - 6}{2} = - 5.\]
\[m^{2} + 7m + 10 =\]
\[= (m + 2)(m + 5)\]
\[D = 49 - 40 = 9\]
\[m_{1} = \frac{- 7 + 3}{2} = - 2,\ \ \]
\[= m_{2} = \frac{- 7 - 3}{2} = - 5.\ \]
