Решебник по алгебре 8 класс Макарычев ФГОС Задание 1324

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\[\textbf{а)}\ \left\{ \begin{matrix} (2x - y)(2x + y) = 3 \\ 2y - 3(x + y) = - 4\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4x^{2} - y^{2} = 3\ \ \ \ \ \ \ \ \ \ \ \\ 2y - 3x - 3y = - 4 \\ \end{matrix} \right.\ \ \]

\[\ \left\{ \begin{matrix} - y - 3x = - 4 \\ 4x^{2} - y^{2} = 3\ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 4 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x^{2} - (4 - 3x)^{2} = 3 \\ \end{matrix} \right.\ \]

\[4x^{2} - 16 + 24x - 9x^{2} - 3 = 0\]

\[- 5x^{2} + 24x - 19 = 0\]

\[5x^{2} - 24x + 19 = 0\]

\[D_{1} = 144 - 95 = 49\]

\[x_{1} = \frac{12 + 7}{5} = \frac{19}{5} = 3,8;\ \ \ \]

\[x_{2} = \frac{12 - 7}{5} = 1.\]

\[y_{1} = 4 - 3 \cdot 3,8 =\]

\[= 4 - 11,4 = - 7,4;\]

\[y_{2} = 4 - 3 \cdot 1 = 1.\]

\[Ответ:(3,8;\ - 7,4);\ \ (1;1).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2 \cdot (x - y) + y = 5\ \ \ \\ (2x - y)^{2} = 5x + 15 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x^{2} - 4x \cdot (2x - 5) + (2x - 5)^{2} - 5x = 15 \\ \end{matrix} \right.\ \]

\[4x^{2} - 8x^{2} + 20x + 4x^{2} -\]

\[- 20x + 25 - 5x = 15\]

\[- 5x = 15 - 25\]

\[- 5x = - 10\]

\[x = 2.\]

\[y = 2 \cdot 2 - 5 = - 1.\]

\[Ответ:\ \ (2; - 1).\]