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МатематикаРешебник по алгебре 9 класс Макарычев Задание 701
Задание 701
\[\boxed{\text{701}\text{\ (701)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ b_{3} = 225,\ \ b_{4} = - 135,\]
\[b_{5} = 81,\]
\[q = \frac{b_{4}}{b_{3}} = - \frac{135}{225} = - \frac{3}{5} = - 0,6\]
\[b_{3} = b_{1} \cdot q^{2}\]
\[b_{1} = \frac{b_{3}}{q^{2}} = 225\ :\left( - \frac{3}{5} \right)^{2} =\]
\[= 225 \cdot \frac{25}{9} = 625\]
\[b_{2} = b_{1} \cdot q = 625 \cdot \left( - \frac{3}{5} \right) =\]
\[= - 375\]
\[b_{6} = b_{5} \cdot q = 81 \cdot \left( - \frac{3}{5} \right) =\]
\[= - \frac{243}{5} = - 48,6.\]
\[\textbf{б)}\ b_{4} = 36,\ \ b_{5} = 54,\]
\[q = \frac{b_{5}}{b_{4}} = \frac{54}{36} = 1,5\]
\[b_{3} = \frac{b_{4}}{q} = 36\ :\frac{3}{2} = 24\]
\[b_{2} = \frac{b_{3}}{q} = 24\ :\frac{3}{2} = 16\]
\[b_{1} = \frac{b_{2}}{q} = 16\ :\frac{3}{2} = \frac{32}{3} = 10\frac{2}{3}.\]
\[\boxed{\text{701.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ (a + 2b)(a - 2b)\left( a^{2} + 4b^{2} \right) =\]
\[= a^{4} - 16b^{4}\]
\[(a + 2b)(a - 2b)\left( a^{2} + 4b^{2} \right) =\]
\[= a^{4} - 16b^{4}\]
\[\left( a^{2} - 4b^{2} \right)\left( a^{2} + 4b^{2} \right) =\]
\[= a^{4} - 16b^{4}\]
\[a^{4} - 16b^{4} = a^{4} - 16b^{4} \Longrightarrow\]
\[\Longrightarrow ч.т.д.\]
\[\textbf{б)}\ (x - 1)(x + 1)\left( x^{2} + 1 \right)\left( x^{4} + 1 \right) =\]
\[= x^{8} - 1\]
\[(x - 1)(x + 1)\left( x^{2} + 1 \right)\left( x^{4} + 1 \right) =\]
\[= x^{8} - 1\]
\[\left( x^{2} - 1 \right)\left( x^{2} + 1 \right)\left( x^{4} + 1 \right) =\]
\[= x^{8} - 1\]
\[\left( x^{4} - 1 \right)\left( x^{4} + 1 \right) = x^{8} - 1\]
\[x^{8} - 1 = x^{8} - 1 \Longrightarrow ч.т.д.\]
\[\left( a^{3} - 8 \right)\left( a^{3} + 8 \right) = a^{6} - 64\]
\[a^{6} - 64 = a^{6} - 64 \Longrightarrow ч.т.д.\]
\[\textbf{г)}\ \left( c^{2} - c - 2 \right)\left( c^{2} + c - 2 \right) =\]
\[= c^{4} - 5c^{2} + 4\]
\[\left( c^{2} - c - 2 \right)\left( c^{2} + c - 2 \right) =\]
\[= c^{4} - 5c^{2} + 4\]
\[\left( \left( c^{2} - 2 \right) - c \right)\left( \left( c^{2} - 2 \right) + c \right) =\]
\[= c^{4} - 5c^{2} + 4\]
\[\left( c^{2} - 2 \right)^{2} - c^{2} =\]
\[= c^{4} - 4c^{2} + 4 - c^{2} =\]
\[= c^{4} - 5c^{2} + 4\]
\[c^{4} - 5c^{2} + 4 = c^{4} - 5c^{2} + 4 \Longrightarrow\]
\[\Longrightarrow ч.т.д.\]
