Изобразите схематически график функции, рассмотренной в задании 8.

Ответ:

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

\[\frac{2}{5c - c^{2}}\]

\[ООФ:\]

\[5c - c^{2} \neq 0\]

\[c(5 - c) \neq 0\]

\[c \neq 0;\ \ c \neq 5.\]

\[c \in ( - \infty;0) \cup (0;5) \cup (5; + \infty).\]

\[\frac{5c}{c^{2} + 1}\]

\[ООФ:\]

\[c^{2} + 1 \neq 0\]

\[c^{2} \neq - 1\]

\[c - любое\ число.\]

\[c \in ( - \infty;\ + \infty).\]

\[\left( \frac{x^{\backslash y + x}}{x - y} - \frac{x^{\backslash x - y}}{y + x} \right)\ :\frac{x^{2}}{x + y} =\]

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

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

\[(5x + 8)\left( 9 - x^{2} \right) = 0\]

\[1)\ 5x + 8 = 0\]

\[5x = - 8\]

\[x = - 1,6.\]

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

\[x^{2} = 9\]

\[x = \pm 3.\]

\[Ответ:x = \pm 3;x = - 1,6.\]

\[x^{4} - 2x^{2} - 8 = 0\]

\[Пусть\ x^{2} = y:\]

\[y^{2} - 2y - 8 = 0\]

\[D_{1} = 1 + 8 = 9\]

\[y_{1} = 1 + 3 = 4;\ \ y_{2} = 1 - 3 = - 2.\]

\[Подставим:\]

\[1)\ x^{2} = 4\]

\[x = \pm 2.\]

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

\[нет\ корней.\]

\[Ответ:x = \pm 2.\]

\[\frac{5}{n} + \frac{4}{n - 3} = 3\ \ \ | \cdot n(n - 3)\]

\[5 \cdot (n - 3) + 4n = 3n(n - 3)\]

\[5n - 15 + 4n = 3n^{2} - 9n\]

\[3n^{2} - 9n - 9n + 15 = 0\ \ \ |\ :3\]

\[n^{2} - 6n + 5 = 0\]

\[D_{1} = 9 - 5 = 4\]

\[n_{1} = 3 + 2 = 5;\ \ n_{2} = 3 - 2 = 1.\]

\[Ответ:при\ n = 1;n = 5.\]

\[Пусть\ x\ \frac{км}{ч} - собственная\ скорость\ лодки.\]

\[\frac{36}{x + 2}\ ч - время\ по\ течению;\]

\[\frac{20}{x - 2}\ ч - время\ против\ течения.\]

\[Составим\ уравнение:\]

\[\frac{36}{x + 2} = \frac{20}{x - 2}\ \]

\[36 \cdot (x - 2) = 20 \cdot (x + 2)\]

\[36x - 72 = 20x + 40\]

\[16x = 112\]

\[x = 7\ \left( \frac{км}{ч} \right) - собственная\ скорость\]

\[лодки.\]

\[Ответ:7\ \frac{км}{ч}.\]

\[\frac{2x - 5}{2x^{2} - 3x - 5} = \frac{2x - 5}{(2x - 5)(x + 1)} = \frac{1}{x + 1}.\]

\[2x^{2} - 3x - 5 = 2 \cdot \left( x - \frac{5}{2} \right)(x + 1) =\]

\[= (2x - 5)(x + 1)\]

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2};\]

\[x_{2} = \frac{3 - 7}{4} = - 1.\]

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

\[y = \frac{1}{x};\ \ \ x \neq 0;\ \ x \neq 1.\]

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

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

\[= (x - 3)\left( 1 - x^{2} \right)\]

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

\[Ответ:x = \pm 1;x = 3\]