Решебник по алгебре 9 класс Макарычев Задание 434

 

\[\boxed{\text{434\ (434).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} 2xy - y = 7 \\ x - 5y = 2 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y(5y + 2) - y = 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 10y^{2}\ + 3y - 7 = 0 \\ \end{matrix} \right.\ \]

\[10y^{2} + 3y - 7 = 0\]

\[D = 9 + 280 = 289\]

\[y_{1,2} = \frac{- 3 \pm 17}{20} = - 1;0,7.\]

\[1)\ y_{1} = - 1;\ \ \ x_{1} = - 3;\]

\[2)\ y_{2} = 0,7;\text{\ \ }x_{2} = 5,5.\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - xy = 33 \\ 4x - y = 17\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 4x - 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - x(4x - 17) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 4x - 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 4x^{2} + 17x - 33 = 0 \\ \end{matrix} \right.\ \]

\[- 2x^{2} + 17x - 33 = 0\]

\[2x^{2} - 17x + 33 = 0\]

\[D = 289 - 8 \cdot 33 = 25\]

\[x_{1,2} = \frac{17 \pm 5}{4} = 3;5,5.\]

\[1)\ x_{1} = 3;\ \ y_{1} = - 5;\]

\[2)\ x_{2} = 5,5;\ \ y_{2} = 5.\]

\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + 2y = 18 \\ 3x = 2y\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3}\text{y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{4}{9}y^{2} + 2y - 18 = 0 \\ \end{matrix} \right.\ \]

\[4y^{2} + 18y - 162 = 0\]

\[2y^{2} + 9y - 81 = 0\]

\[D = 81 + 4 \cdot 2 \cdot 81 = 729 = 27^{2}\]

\[y_{1,2} = \frac{- 9 \pm 27}{4} = 4,5;\ - 9.\]

\[1)\ y_{1} = 4,5;\ \ x_{1} = 3;\]

\[2)\ y_{2} = - 9;\ \ x_{2} = - 6.\]

\[\textbf{г)}\ \left\{ \begin{matrix} x - y - 4 = 0 \\ x^{2} + y^{2} = 8,5\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 8y + 16 + y^{2} = 8,5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y^{2} + 8y + 7,5 = 0 \\ \end{matrix} \right.\ \]

\[2y^{2} + 8y + 7,5 = 0\]

\[D_{1} = 16 - 7,5 \cdot 2 = 1\]

\[y_{1,2} = \frac{- 4 \pm 1}{2} = - 1,5;\ - 2,5.\]

\[1)\ y_{1} = - 1,5;\ \ x_{1} = 2,5;\]

\[2)\ y_{2} = - 2,5;\ \ x_{2} = 1,5.\]

\[\textbf{д)}\ \left\{ \begin{matrix} x^{2} + 4y = 10 \\ x - 2y = - 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 20y + 25 + 4y = 10 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 16y + 15 = 0 \\ \end{matrix} \right.\ \]

\[4y^{2} - 16y + 15 = 0\]

\[D_{1} = 64 - 60 = 4\]

\[y_{1,2} = \frac{8 \pm 2}{4} = 1,5;2,5.\]

\[1)\ y_{1} = 1,5;\ \ x_{1} = - 2;\]

\[2)\ y_{2} = 2,5;\ \ x_{2} = 0.\]

\[\textbf{е)}\ \left\{ \begin{matrix} x - 2y + 1 = 0 \\ 5xy + y^{2} = 16\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y(2y - 1) + y^{2} = 16 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 11y^{2} - 5y - 16 = 0 \\ \end{matrix} \right.\ \]

\[D = 25 + 11 \cdot 16 \cdot 4 = 729\]

\[y_{1,2} = \frac{5 \pm 27}{22} = - 1;1\frac{5}{11}.\]

\[1)\ y_{1} = - 1;\ \ x_{1} = - 3;\]

\[2)\ y_{2} = 1\frac{5}{11};\ \ x_{2} = 1\frac{10}{11}.\]

\[Ответ:а)\ ( - 3;\ - 1);\ \ (5,5;0,7);\ \ \]

\[\textbf{б)}\ \ (3;\ - 5);(5,5;5);\]

\[\textbf{в)}\ (3;4,5);\ \ ( - 6;\ - 9);\ \ \]

\[\textbf{г)}\ (2,5;\ - 1,5);\ \ (1,5;\ - 2,5);\]

\[\textbf{д)}\ ( - 2;1,5);\ \ (0;2,5);\ \ \]

\[\textbf{е)}\ ( - 3;\ - 1);\ \ \left( 1\frac{10}{11};1\frac{5}{11} \right).\]

\(\boxed{\text{434.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

\[Пусть\ x - площадь\ опоры;\]

\[\frac{30}{x}\ \frac{кг}{дм^{2}} - давление\ груза;\]

\[\frac{30 - 2}{x - 1} =\]

\[= \frac{28}{x - 1}\frac{кг}{дм^{2}} - давление,\ \]

\[которое\ на\ 1\ \frac{кг}{дм^{2}}\ \]

\[больше\ реального.\]

\[Составим\ уравнение:\]

\[\frac{28}{x - 1} = \frac{30}{x} - 1\]

\[28x = 30 \cdot (x - 1) - x(x - 1)\]

\[28x = 30x - 30 - x^{2} + x\]

\[x^{2} + x - 30 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 5;\ \ \ \ x_{2} = - 6.\]

\[Так\ как\ \ x > 0 \Longrightarrow x = 5.\]

\[Ответ:5\ дм^{2}.\]