ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1014

\[\boxed{\mathbf{1014}\mathbf{.}}\]

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

\[(x + 1)^{2} = 1 - x\]

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

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

\[x \bullet (x + 3) = 0\]

\[x_{1} = 0\ и\ x_{2} = - 3.\]

\[\textbf{а)}\ (x + 1)^{2} > 0\]

\[x + 1 \neq 0\]

\[x \neq - 1.\]

\[\textbf{б)}\ 1 - x > 0\ \]

\[x < 1.\]

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

\[= \frac{2}{6} + 1 - \frac{3}{6} = 1 - \frac{1}{6} = \frac{5}{6}.\]

\[Ответ:\ \ \frac{5}{6}.\]

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

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

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

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

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

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

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

\[\textbf{а)}\ 4 - x^{2} > 0\]

\[x^{2} < 4\]

\[- 2 < x < 2.\]

\[\textbf{б)}\ x + 2 > 0\]

\[x > - 2.\]

\[Ответ:\ \ 6\frac{1}{6}.\]

\[3)\ y = 4x - x^{2}\text{\ \ }и\ \ y = 4 - x\]

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

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

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

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

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

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

\[\textbf{а)}\ 4x - x^{2} > 0\]

\[x \bullet (1 - x) > 0\]

\[0 < x < 1.\]

\[\textbf{б)}\ 4 - x > 0\ \]

\[x < 4.\]

\[= \left. \ \left( 2x^{2} - \frac{x^{3}}{3} \right) \right|_{0}^{1} + \left. \ \left( 4x - \frac{x^{2}}{2} \right) \right|_{1}^{4} =\]

\[= 2 - \frac{1}{3} + 16 - 8 - 4 + \frac{1}{2} =\]

\[= 6 - \frac{1}{3} + \frac{1}{2} = \frac{36 - 2 + 3}{6} =\]

\[= \frac{37}{6} = 6\frac{1}{6}.\]

\[Ответ:\ \ 6\frac{1}{6}.\]

\[4)\ y = 3x^{2}\text{\ \ }и\ \ y = 1,5x + 4,5\]

\[3x^{2} = 1,5x + 4,5\]

\[6x^{2} - 3x - 9 = 0\]

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

\[D = 1^{2} + 4 \bullet 2 \bullet 3 = 1 + 24 = 25\]

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

\[x_{2} = \frac{1 + 5}{2 \bullet 2} = \frac{6}{4} = 1,5.\]

\[\textbf{а)}\ 3x^{2} > 0\]

\[3x^{2} \neq 0\ \]

\[x \neq 0.\]

\[\textbf{б)}\ 1,5x + 4,5 > 0\]

\[1,5x > - 4,5\ \]

\[x > - 3.\]

\[= \left. \ x^{3}\ \right|_{- 1}^{0} + \left. \ \left( \frac{3}{4}x^{2} + \frac{9}{2}x \right) \right|_{- 3}^{- 1} =\]

\[= 1 + \frac{3}{4} - \frac{9}{2} - \frac{27}{4} + \frac{27}{2} =\]

\[= 1 - \frac{24}{4} + \frac{18}{2} = 1 - 6 + 9 = 4.\]

\[Ответ:\ \ 4.\]