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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 772
Задание 772
\[\boxed{\text{772.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[x^{2} + px + 90 = 0,\ \ \]
\[\left( x_{2} - x_{1} \right)^{2} = 81\]
| \[\left\{ \begin{matrix} x_{1} + x_{2} = - p \\ x_{1}x_{2} = 90\ \ \ \ \ \ \\ \end{matrix} \right.\ \ \ \ \Longrightarrow \ \] |
\[x_{2}^{2} - 2x_{1}x_{2} + x_{1}^{2} = 81\] \[x_{1}^{2} + x_{2}^{2} - 2x_{1}x_{2} = 81\ \] |
|---|---|
| \[\left\{ \begin{matrix} x_{1} + x_{2} = - p \\ x_{1} = \frac{90}{x_{2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \ \ \ \ \Longrightarrow \ \] |
\[x_{1}^{2} + x_{2}^{2} - 2 \cdot 90 = 81\] \[x_{1}^{2} + x_{2}^{2} = 261\ \] |
|
\[p = - \left( x_{1} + x_{2} \right)\] \[p_{1} = - (15 + 6) = - 21\] \[p_{2} = - ( - 15 - 6) = 21\] \[p_{3} = - (6 + 15) = - 21\] \[p_{4} = - ( - 6 - 15) = 21\] \[Ответ:p = \pm 21.\ \] |
\[\left( \frac{90}{x_{2}} \right)^{2} + x_{2}^{2} = 261\] \[\frac{8100}{x_{2}^{2}} + x_{2}^{2} = 261\ \ \ \ | \cdot x_{2}^{2}\] \[x_{2}^{4} - 261x_{2}^{2} + 8100 = 0\] \[D = 68121 - 32400 = 35721 = 189^{2}\] \[x_{1,2}^{2} = \frac{261 \pm 189}{2} = \frac{450}{2};\frac{72}{2}\] \[x_{1}^{2} = 225;\ \ \ \ \ \ \ \ \ \ \ \ x_{2}^{2} = 36\] \[x_{1} = \pm 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2} = \pm 6\] \[x_{1} = \frac{90}{x_{2}} = \frac{90}{\pm 15}\text{\ \ \ \ \ }x_{1} = \frac{90}{x_{2}} = \frac{90}{\pm 6}\] \[x_{1} = \pm 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{1} = \pm 15\ \ \ \ \] |
