Решебник по алгебре 9 класс Макарычев Задание 709

\[\boxed{\text{709\ (709).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[b_{n} = b_{1} \cdot q^{n - 1},\]

\[b_{m} = b_{1} \cdot q^{m - 1},\]

\[b_{1} = \frac{b_{m}}{q^{m - 1}},\]

\[b_{n} = \frac{b_{m}}{q^{m - 1}} \cdot q^{n - 1} =\]

\[= b_{m} \cdot q^{n - 1 - m + 1} = b_{m} \cdot q^{n - m}.\]

\[\boxed{\text{709.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x^{2} - 4x}{x^{2} + 7x}\ :\frac{24 - 6x}{49 - x^{2}} =\]

\[= \frac{x(x - 4)}{x(x + 7)} \cdot \frac{(7 - x)(7 + x)}{6 \cdot (4 - x)} =\]

\[= - \frac{7 - x}{6} = \frac{x - 7}{6}\]

\[\textbf{б)}\ \frac{y^{3} - 16y}{2y + 18}\ :\frac{4 - y}{y^{2} + 9y} =\]

\[= \frac{y(y - 4)(y + 4)}{2 \cdot (y + 9)} \cdot \frac{y(y + 9)}{4 - y} =\]

\[= - \frac{y^{2}(y + 4)}{2} = - \frac{y³ + 4y²}{2}\]

\[\textbf{в)}\ \frac{(a + b)^{2} - 2ab}{4a^{2}}\ :\frac{a^{2} + b^{2}}{\text{ab}} =\]

\[= \frac{a^{2} + 2ab + b^{2} - 2ab}{4a^{2}} \cdot \frac{\text{ab}}{a^{2} + b^{2}} =\]

\[= \frac{b}{4a}\]

\[\textbf{г)}\ \frac{5c^{3} - 5}{c + 2}\ :\frac{(c + 1)^{2} - c}{13c + 26} =\]

\[= \frac{5 \cdot \left( c^{3} - 1 \right)}{c + 2} \cdot \frac{13 \cdot (c + 2)}{c^{2} + 2c + 1 - c} =\]

\[= \frac{65 \cdot (c - 1)\left( c^{2} + c + 1 \right)}{c^{2} + c + 1} =\]

\[= 65c - 65.\]