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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 689
Задание 689
\[\boxed{\mathbf{689\ (689).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\left\{ \begin{matrix} 2x - y = 13\ \ \\ x^{2} - y^{2} = 23 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 2x - 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - (2x - 13)^{2} = 23 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x^{2} + 52x - 169 - 23 = 0\]
\[- 3x^{2} + 52x - 192 = 0\]
\[3x^{2} - 52x + 192 = 0\]
\[D = 2704 - 2304 = 400\]
\[x_{1} = \frac{52 - 20}{6} = \frac{16}{3}\]
\[x_{2} = \frac{52 + 20}{6} = 12\]
\[\left\{ \begin{matrix} x = \frac{16}{3}\text{\ \ \ } \\ y = - \frac{7}{3}\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 12 \\ y = 11 \\ \end{matrix} \right.\ \]
\[Ответ:\left( \frac{16}{3};\ - \frac{7}{3} \right);\ \ (12;11).\]
\[2)\ \left\{ \begin{matrix} 2x^{2} - y^{2} = 23 \\ 2x^{2} + y^{2} = 41 \\ \end{matrix} \right.\ \text{\ \ }( + )\text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} 4x^{2} = 64\ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - y^{2} = 23 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x^{2} = 16\ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - y^{2} = 23 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = \pm 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y² = 2 \cdot 16 - 23 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \pm 4 \\ y = \pm 3 \\ \end{matrix} \right.\ \]
\[{\left\{ \begin{matrix} x = 4 \\ y = 3\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 4\ \ \ \\ y = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ } }{или\ \ \ \ \ \ \left\{ \begin{matrix} x = - 4 \\ y = 3\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }или\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = - 4 \\ y = - 3 \\ \end{matrix} \right.\ \]
\[Ответ:(4;3);\ \ (4;\ - 3);\ \ \]
\[( - 4;3);\ \ ( - 4;\ - 3).\]
\[\boxed{\mathbf{689.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\left\{ \begin{matrix} b_{4} - b_{2} = 30 \\ b_{4} - b_{3} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1}q^{3} - b_{1}q = 30 \\ b_{1}q^{3} - b_{1}q^{2} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1}q\left( q^{2} - 1 \right) = 30 \\ b_{1}q^{2}(q - 1) = 24 \\ \end{matrix} \right.\ \ \ \ \ |\ :\]
\[\frac{b_{1}q\left( q^{2} - 1 \right)}{b_{1}q^{2}(q - 1)} = \frac{30}{24}\text{\ \ }\]
\[\frac{(q - 1)(q + 1)}{q(q - 1)} = \frac{5}{4}\text{\ \ }\]
\[\frac{q + 1}{q} = \frac{5}{4}\]
\[5q = 4q + 4 \Longrightarrow \ \ q = 4.\]
\[b_{1} \cdot 4 \cdot \left( 4^{2} - 1 \right) = 30\ \ \]
\[b_{1} \cdot 4 \cdot 15 = 30\ \ \]
\[b_{1} = \frac{30}{60} \Longrightarrow \ \ b_{1} = 0,5.\]
\[Ответ:\ b_{1} = 0,5;\ \ \ q = 4.\]
\[2)\ \left\{ \begin{matrix} b_{2} - b_{5} = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{3} + b_{4} + b_{5} = - 117 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1}q - b_{1}q^{4} = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2} + b_{1}q^{3} + b_{1}q^{4} = - 117 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} b_{1}q\left( 1 - q^{3} \right) = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q + q^{2} \right) = - 117 \\ \end{matrix} \right.\ \ \ \ |\ :\]
\[\frac{b_{1}q\left( 1 - q^{3} \right)}{b_{1}q^{2}\left( 1 + q + q^{2} \right)} = \frac{78}{- 117}\]
\[\frac{(1 - q)\left( 1 + q + q^{2} \right)}{q\left( 1 + q + q^{2} \right)} = - \frac{2}{3}\text{\ \ }\]
\[\frac{1 - q}{q} = - \frac{2}{3}\]
\[3 \cdot (1 - q) = - 2q\ \ \]
\[3 - 3q = - 2q \Longrightarrow \ \ q = 3.\]
\[b_{1} \cdot 3 \cdot \left( 1 - 3^{3} \right) = 78\ \ \]
\[b_{1} \cdot 3 \cdot ( - 26) = 78\ \]
\[b_{1} = - \frac{78}{78} = - 1.\]
\[Ответ:\ b_{1} = - 1;\ q = 3.\]
