ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 823

\[\boxed{\mathbf{823}\mathbf{.}}\]

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

\[f^{'}(x) =\]

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

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

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

\[= \frac{3}{(x + 1)^{2}}\]

\[\frac{3}{(x + 1)^{2}} = 3\]

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

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

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

\[x(x + 2) = 0\]

\[x_{1} = 0\ \text{\ \ }x_{2} = - 2.\]

\[Ответ:\ \ - 2\ \ 0.\]

\[\boxed{\mathbf{824}\mathbf{.}}\]

\[f(x) = (x - 1)(x - 2)(x - 3) = \ \]

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

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

\[= x^{3} - 6x^{2} + 11x - 6\]

\[f^{'}(x) =\]

\[= \left( x^{3} \right)^{'} - 6 \bullet \left( x^{2} \right)^{'} + (11x - 6)^{'} =\]

\[= 3x^{2} - 6 \bullet 2x + 11 =\]

\[= 3x^{2} - 12x + 11\]

\[3x^{2} - 12x + 11 = 11\]

\[3x^{2} - 12x = 0\]

\[3x \bullet (x - 4) = 0\]

\[x_{1} = 0\text{\ \ }x_{2} = 4.\]

\[Ответ:\ \ 0\ \ 4.\]