ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 606

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\[1)\ \left\{ \begin{matrix} \sqrt{x} + \sqrt{y} = 5 \\ x - y = 10\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \sqrt{x} + \sqrt{y} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( \sqrt{x} + \sqrt{y} \right)\left( \sqrt{x} - \sqrt{y} \right) = 10 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sqrt{x} + \sqrt{y} = 5\ \ \ \ \ \ \ \ \ \ \ \ \\ 5 \cdot \left( \sqrt{x} - \sqrt{y} \right) = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

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

\[2\sqrt{x} = 7\]

\[\sqrt{x} = 3,5\]

\[x = 12,25.\]

\[\sqrt{y} = 5 - \sqrt{x} = 5 - 3,5 = 1,5;\]

\[y = 2,25.\]

\[Ответ:(12,25;2,25).\]

\[2)\ \left\{ \begin{matrix} \sqrt{x} - \sqrt{y} = 4 \\ x - y = 24\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \sqrt{x} - \sqrt{y} = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( \sqrt{x} - \sqrt{y} \right)\left( \sqrt{x} + \sqrt{y} \right) = 24 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sqrt{x} - \sqrt{y} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ 4 \cdot \left( \sqrt{x} + \sqrt{y} \right) = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} \sqrt{x} - \sqrt{y} = 4 \\ \sqrt{x} + \sqrt{y} = 6 \\ \end{matrix} \right.\ ( + )\]

\[2\sqrt{x} = 10\]

\[\sqrt{x} = 5\]

\[x = 25.\]

\[\sqrt{y} = 6 - \sqrt{x} = 6 - 5 = 1\]

\[y = 1.\]

\[Ответ:(25;1).\]

\[3)\ \left\{ \begin{matrix} \sqrt{x + 1} - \sqrt{y - 1} = 1 \\ x - y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} y = x - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{x + 1} - \sqrt{x - 3 - 1} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\sqrt{x + 1} - \sqrt{x - 4} = 1\]

\[\left( \sqrt{x + 1} - \sqrt{x - 4} \right)^{2} = 1\]

\[x + 1 - 2\sqrt{(x + 1)(x - 4)} +\]

\[+ x - 4 = 1\]

\[2\sqrt{x^{2} - 3x - 4} = 2x - 4\ \ \ \ \ |\ :2\]

\[\sqrt{x^{2} - 3x - 4} = x - 2\]

\[x^{2} - 3x - 4 = (x - 2)^{2}\]

\[x^{2} - 3x - 4 = x^{2} - 4x + 4\]

\[x = 8.\]

\[y = 8 - 3 = 5.\]

\[Ответ:(8;5).\]

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

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

\[\left( \sqrt{x + 2} + \sqrt{x + 1} \right)^{2} = 3^{2}\]

\[x + 2 + 2\sqrt{(x + 2)(x + 1)} +\]

\[+ x + 1 = 9\]

\[2\sqrt{x^{2} + 3x + 2} = 6 - 2x\ \ \ \ |\ :2\]

\[\sqrt{x^{2} + 3x + 2} = 3 - x\]

\[x^{2} + 3x + 2 = (3 - x)^{2}\]

\[x^{2} + 3x + 2 = 9 - 6x + x^{2}\]

\[9x = 7\]

\[x = \frac{7}{9}.\]

\[y = 2 - \frac{7}{9} = 1\frac{2}{9}.\]

\[Ответ:\ \ \left( \frac{7}{9};1\frac{2}{9} \right).\]