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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 638
Задание 638
\[\boxed{\text{638\ (638).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x^{2} - 10x + q = 0;\ \ \text{\ \ }x_{1} - x_{2} = 6\]
\[по\ теореме\ Виета:\]
\[x_{1} + x_{2} = 10;\ \ \ x_{1}x_{2} = q,\ тогда:\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = 10 \\ x_{1} - x_{2} = 6 \\ \end{matrix} + \right.\ \]
\[x_{1} + x_{2} + x_{1} - x_{2} = 10 + 6\]
\[2x_{1} = 16\]
\[x_{1} = 8\]
\[x_{1} + x_{2} = 10\]
\[x_{2} = 10 - 8 = 2\]
\[q = x_{1}x_{2} = 8 \cdot 2 = 16\]
\[Ответ:q = 16.\ \ \]
\[\boxed{\text{638.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ \frac{5}{y - 2} - \frac{y}{y - 3} =\]
\[= \frac{1}{y}\ \ \ \ \ \ | \cdot y(y - 2)(y - 3)\]
\[y \neq 0\]
\[y - 2 \neq 0,\ \ y \neq 2\]
\[y - 3 \neq 0,\ \ y \neq 3\]
\[5y \cdot (y - 3) - 4y \cdot (y - 2) =\]
\[= (y - 2)(y - 3)\]
\[5y^{2} - 15y - 4y^{2} + 8y =\]
\[= y^{2} - 3y - 2y + 6\]
\[- 2y = 6\]
\[y = - 3\]
\[Ответ:y = - 3.\]
\[\textbf{б)}\frac{1}{2 \cdot (x + 1)} + \frac{1}{x + 2} =\]
\[= \frac{3}{x + 3}\text{\ \ \ \ \ \ }\]
\[| \cdot 2(x + 1)(x + 2)(x + 3)\]
\[x + 1 \neq 0,\ \ x \neq - 1\]
\[x + 2 \neq 0,\ \ x \neq - 2\]
\[x + 3 \neq 0,\ \ x \neq - 3\ \]
\[(x + 2)(x + 3) +\]
\[+ 2 \cdot (x + 1)(x + 3) =\]
\[= 6 \cdot (x + 1)(x + 2)\]
\[x^{2} + 3x + 2x + 6 + 2x^{2} + 6x +\]
\[+ 2x + 6 = 6x^{2} +\]
\[+ 12x + 6x + 12\]
\[- 3x^{2} - 5x = 0\]
\[- x(3x + 5) = 0\]
\[x = 0\ \ \ \ \ \ \ \ \ \ \ \ 3x + 5 = 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{5}{3}\]
\[Ответ:x = - 1\frac{2}{3};\ \ x = 0.\]
\[\textbf{в)}\ \frac{1}{x + 2} + \frac{1}{x^{2} - 2x} = \frac{8}{x^{3} - 4x}\ \]
\[\frac{1}{x + 2} + \frac{1}{x(x - 2)} =\]
\[= \frac{8}{x(x - 2)(x + 2)}\text{\ \ \ \ \ \ \ \ }\]
\[| \cdot x(x - 2)(x + 2)\]
\[x \neq 0\]
\[x - 2 \neq 0,\ \ x \neq 2\]
\[x + 2 \neq 0,\ \ x \neq - 2\]
\[x(x - 2) + x + 2 = 8\]
\[x^{2} - 2x + x + 2 = 8\]
\[x^{2} - x - 6 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1,2} = \frac{1 \pm 5}{2}\]
\[x_{1} = 3;\ \ x_{2} = - 2 - не\ \]
\[принадлежит\ по\ ОДЗ\]
\[Ответ:x = 3.\]
\[\textbf{г)}\frac{10}{y^{3} - y} + \frac{1}{y - y^{2}} = \frac{1}{1 + y}\]
\[\frac{10}{y(y - 1)(y + 1)} +\]
\[+ \frac{1}{y(1 - y)} = \frac{1}{1 + y}\]
\[\frac{10}{y(y - 1)(y + 1)} - \frac{1}{y(y - 1)} =\]
\[= \frac{1}{y + 1}\ \ \ \ \ | \cdot y(y - 1)(y + 1)\]
\[y \neq 0\]
\[y - 1 \neq 0,\ \ y \neq 1\]
\[y + 1 \neq 0,\ \ y \neq - 1\]
\[10 - (y + 1) = y(y - 1)\]
\[10 - y - 1 = y^{2} - y\]
\[y^{2} = 9\]
\[y = \pm 3\]
\[Ответ:y = \pm 3.\]
\[\textbf{д)}\ 1 + \frac{45}{x^{2} - 8x + 16} = \frac{14}{x - 4}\]
\[x - 4 \neq 0,\ \ x \neq 4\]
\[1 + \frac{45}{(x - 4)^{2}} = \frac{14}{x - 4}\ | \cdot (x - 4)^{2}\]
\[(x - 4)^{2} + 45 = 14 \cdot (x - 4)\]
\[x^{2} - 8x + 16 + 45 = 14x - 56\]
\[x^{2} - 22x + 117 = 0\]
\[D = 484 - 468 = 16\]
\[x_{1,2} = \frac{22 \pm 4}{2}\]
\[x_{1} = 13;\ \ x_{2} = 9\]
\[Ответ:x = \left\{ 9;13 \right\}.\]
\[\textbf{е)}\frac{5}{x - 1} - \frac{4}{3 - 6x + 3x^{2}} = 3\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[\frac{5}{x - 1} - \frac{4}{3 \cdot \left( 1 - 2x + x^{2} \right)} = 3\]
\[\frac{5}{x - 1} - \frac{4}{3 \cdot (1 - x)^{2}} = 3\]
\[\frac{5}{x - 1} - \frac{4}{3 \cdot (x - 1)^{2}} =\]
\[= 3\ \ \ \ \ \ \ \ | \cdot 3 \cdot (x - 1)^{2}\]
\[15 \cdot (x - 1) - 4 = 9 \cdot (x - 1)^{2}\]
\[15x - 15 - 4 =\]
\[= 9 \cdot \left( x^{2} - 2x + 1 \right)\]
\[15x - 19 = 9x^{2} - 18x + 9\]
\[9x^{2} - 33x + 28 = 0\]
\[D = 1089 - 1008 = 81\]
\[x_{1,2} = \frac{33 \pm 9}{18} = \frac{42}{18};\ \frac{24}{18}\]
\[x_{1} = \frac{7}{3};\ \ x_{2} = \frac{4}{3}\]
\[Ответ:x = 1\frac{1}{3};\ \ x = 2\frac{1}{3}\text{.\ }\]
