Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1020

\[\boxed{\mathbf{1020}\mathbf{.}}\]

\[1)\ y = 6x - x^{2}\text{\ \ }и\ \ y = x + 4\]

\[6x - x^{2} = x + 4\]

\[x^{2} + x - 6x + 4 = 0\]

\[x^{2} - 5x + 4 = 0\]

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2} = 1\ \ и\ \ \]

\[x_{2} = \frac{5 + 3}{2} = 4.\]

\[= \left. \ \left( 3x^{2} - \frac{x^{3}}{3} - \frac{x^{2}}{2} - 4x \right) \right|_{1}^{4} =\]

\[= 33 - 21 - 7,5 = 4,5.\]

\[Ответ:\ \ 4,5.\]

\[2)\ y = 4 - x^{2}\text{\ \ }и\ \ y = x + 2\]

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

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

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\ \ \]

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

\[= 2 - \frac{1}{3} - \frac{1}{2} + 4 - \frac{8}{3} + \frac{4}{2} =\]

\[= 6 - 3 + 1,5 = 4,5.\]

\[Ответ:\ \ 4,5.\]