Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1018

\[\boxed{\mathbf{1018}\mathbf{.}}\]

\[1)\ y = 6x^{2}\text{\ \ }и\ \ y = (x - 3)(x - 4)\]

\[6x^{2} = (x - 3)(x - 4)\]

\[6x^{2} = x^{2} - 4x - 3x + 12\]

\[6x^{2} = x^{2} - 7x + 12\]

\[5x^{2} + 7x - 12 = 0\]

\[D = 7^{2} + 4 \bullet 5 \bullet 12 =\]

\[= 49 + 240 = 289\]

\[x_{1} = \frac{- 7 - 17}{2 \bullet 5} = - 2,4\ \ и\ \ \]

\[x_{2} = \frac{- 7 + 17}{2 \bullet 5} = 1.\]

\[\textbf{а)}\ 6x^{2} > 0\]

\[6x^{2} \neq 0\]

\[x \neq 0.\]

\[\textbf{б)}\ (x - 3)(x - 4) > 0\]

\[x < 3\ или\ x > 4.\]

\[= 35 - \frac{56}{2} - \frac{1}{3} = 35 - 28 - \frac{1}{3} =\]

\[= 7 - \frac{1}{3} = 6\frac{2}{3}.\]

\[Ответ:\ \ 6\frac{2}{3}.\]

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

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

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

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

\[2x \bullet (x - 2) = 0\]

\[x_{1} = 0\ и\ x_{2} = 2.\]

\[\textbf{а)}\ 4 - x^{2} > 0\]

\[x^{2} < 4\]

\[- 2 < x < 2.\]

\[\textbf{б)}\ (x - 2)^{2} > 0\]

\[(x - 2)^{2} \neq 0\]

\[x \neq 2.\]

\[= - \left( - 8 + \frac{8}{3} \right) + \frac{0^{3}}{3} - \frac{( - 2)^{3}}{3} =\]

\[= 8 - \frac{8}{3} + \frac{8}{3} = 8.\]

\[Ответ:\ \ 8.\]