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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 390
Задание 390
\[\boxed{\text{390\ (390).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left( x^{2} + 17 \right)(x - 6)(x + 2) < 0\]
\[x^{2} + 17 > 0\ при\ любом\ x\]
\[\Longrightarrow (x + 2)(x - 6) < 0\]
\[x \in ( - 2;6).\]
\[\textbf{б)}\ \left( 2x^{2} + 1 \right) \cdot x \cdot (x - 4) > 0\]
\[так\ как\ 2x^{2} + 1 > 0\]
\[\Longrightarrow x(x - 4) > 0\]
\[x \in ( - \infty;0) \cup (4; + \infty).\]
\[\textbf{в)}\ (x - 1)^{2}(x - 24) < 0\]
\[(x - 1)^{2} \geq 0\ при\ любом\ \]
\[значении\ x;\]
\[x - 24 < 0;\ \ \ x \neq 1.\]
\[x \in ( - \infty;1) \cup (1;24).\]
\[\textbf{г)}\ (x + 7)(x - 4)^{2}(x - 21) > 0\]
\[(x - 4)^{2} \geq 0\ при\ любом\ x;\]
\[(x + 7)(x - 21) > 0;\ \ \ \ x \neq 4.\]
\[x \in ( - \infty; - 7) \cup (21; + \infty).\]
\[\boxed{\text{390.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy - y^{2} = 11 \\ x - 2y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} + 4y + 1 + 2y^{2} + y - y^{2} = 11 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y^{2} + 5y - 10 = 0 \\ \end{matrix} \right.\ \]
\[5y^{2} + 5y - 10 = 0\ \ \ \ |\ :5\]
\[y^{2} + y - 2 = 0\]
\[y_{1} + y_{2} = - 1;\ \ \ y_{2} \cdot y_{2} = - 2\]
\[y_{1} = 1;\ \ \ y_{2} = - 2.\]
\[1)\ y_{1} = 1;\ \ x_{1} = 3;\]
\[2)\ y_{2} = - 2;\ \ x_{2} = - 3.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy - 3y = 9 \\ 3x + 2y = - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 1,5x^{2} - 0,5x + 4,5x + 1,5 - 9 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \\ - 0,5x^{2} + 4x - 7,5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[0,5x^{2} - 4x + 7,5 = 0\ \ \ \ \ \ | \cdot 2\]
\[x^{2} - 8x + 15 = 0\]
\[D_{1} = 16 - 15 = 1\]
\[x_{1} = 4 + 1 = 5;\ \ \ \]
\[x_{2} = 4 - 1 = 3.\]
\[1)\ x_{1} = 5;\ \ y_{1} = - 8;\]
\[2)\ x_{2} = 3;\ \ \ y_{2} = - 5.\]
\[Ответ:а)\ \ (3;1);( - 3;\ - 2);\ \ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (5;\ - 8);\ \ (3; - 5).\]
