Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1010

\[1)\ y = x^{2}\text{\ \ }и\ \ y = x + 6\]

\[x^{2} = x + 6\]

\[x^{2} - x - 6 = 0\]

\[D = 1 + 24 = 25\]

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

\[x_{2} = \frac{1 + 5}{2} = 3.\]

\[Ответ:\ \ да.\]

\[2)\ y = \frac{3}{x}\text{\ \ }и\ \ y = 4(x + 1)\]

\[\frac{3}{x} = 4(x + 1)\ \ \ \ \ | \bullet x\]

\[3 = 4x(x + 1)\]

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

\[D = 16 + 48 = 64\]

\[x_{1} = \frac{- 4 - 8}{2 \bullet 4} = - 1,5;\]

\[x_{2} = \frac{- 4 + 8}{2 \bullet 4} = 0,5.\]

\[Ответ:\ \ да.\]

\[3)\ y = \frac{1}{8}x^{2}\text{\ \ }и\ \ y = \frac{1}{x}:\]

\[\frac{1}{8}x^{2} = \frac{1}{x}\ \ \ \ \ | \bullet 8x\]

\[x^{3} = 8\]

\[x = \sqrt[3]{8} = 2.\]

\[Ответ:\ \ да.\]

\[4)\ y = 2x - 1\ \ и\ \ y = \frac{1}{x}\]

\[2x - 1 = \frac{1}{x}\ \ \ \ \ | \bullet x\]

\[2x^{2} - x - 1 = 0\]

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{1 - 3}{2 \bullet 2} = - 0,5;\]

\[x_{2} = \frac{1 + 3}{2 \bullet 2} = 1.\]

\[Ответ:\ \ да.\]