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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 281
Задание 281
\(\boxed{\text{281\ (281).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\)
\[\textbf{а)}\ x^{4} - 47x^{2} - 98\]
\[x^{2} = t:\]
\[t^{2} - 47t - 98 = 0\]
\[t_{1} + t_{2} = 47;\ \ \ t_{1} \cdot t_{2} = - 98\]
\[t_{1} = 49;\ \ \ t_{2} = - 2.\]
\[t^{2} - 47t - 98 =\]
\[= (t + 2)(t - 49) =\]
\[= (t + 2)(t - 49).\]
\[x^{4} - 47x^{2} - 98 =\]
\[= \left( x^{2} + 2 \right)\left( x^{2} - 49 \right) =\]
\[= \left( x^{2} + 2 \right)(x - 7)(x + 7).\]
\[\textbf{б)}\ x^{4} - 85x^{2} + 1764\]
\[x^{2} = t:\]
\[t^{2} - 85t + 1764 = 0\]
\[t_{1} + t_{2} = 85;\ \ \ t_{1} \cdot t_{2} = 1764\]
\[t_{1} = 36;\ \ \ t_{2} = 49.\]
\[t^{2} - 85t + 1764 =\]
\[= (t - 36)(t - 49).\]
\[x^{4} - 85x^{2} + 1764 =\]
\[= \left( x^{2} - 36 \right)\left( x^{2} - 49 \right) =\]
\[= (x - 6)(x + 6)(x - 7)(x + 7).\]
\[\boxed{\text{281.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ y =\]
\[= \sqrt{25 - x^{2}} + \sqrt{9x - x^{2} - 14}\]
\[ОДЗ:\]
\[\left\{ \begin{matrix} 25 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ 9x - x^{2} - 14 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ x^{2} - 9x + 14 \leq 0\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ (x - 2)(x - 7) \leq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 9x + 14 = 0\]
\[x_{1} + x_{2} = 9;\ \ \ x_{1} \cdot x_{2} = 14\]
\[x_{1} = 2;\ \ \ \ \ \ \ \ \ \ \ x_{2} = 7\]
\[\ x \in \lbrack 2;5\rbrack.\]
\[Ответ:\ x = 2;3;4;5.\]
\[\textbf{б)}\ y = \sqrt{8x - x^{2} - 12} +\]
\[+ \sqrt{16 - x^{2}}\]
\[ОДЗ:\]
\[\left\{ \begin{matrix} 8x - x^{2} - 12 \geq 0 \\ 16 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 8x + 12 \leq 0 \\ x^{2} - 16 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 2)(x - 6) \leq 0 \\ (x - 4)(x + 4) \leq 0 \\ \end{matrix} \right.\ \ \]
\[x^{2} - 8x + 12 = 0\]
\[D_{1} = 16 - 12 = 4\]
\[x_{1} = 4 + 2 = 6;\ \ \]
\[x_{2} = 4 - 2 = 2.\]
\[x \in \lbrack 2;4\rbrack.\]
\[Ответ:x = 2;3;4.\]
