Решебник по алгебре 8 класс Макарычев ФГОС Задание 636

 

\[\boxed{\text{636\ (636).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\frac{1}{11 + 2\sqrt{30}} + \frac{1}{11 - 2\sqrt{30}} = 22\]

\[\frac{22}{121 - 4 \cdot 30} = 22\]

\[\frac{22}{121 - 120} = 22\]

\[22 = 22 - что\ и\ требовалось\ \]

\[доказать.\]

\[\textbf{б)}\frac{\sqrt{5} + 2}{\sqrt{5} - 2} + \frac{\sqrt{5} - 2}{\sqrt{5} + 2} = 18\]

\[\frac{5 + 4\sqrt{5} + 4 + 5 - 4\sqrt{5} + 4}{5 - 4} = 18\]

\[\frac{18}{1} = 18\]

\[18 = 18 - что\ и\ требовалось\ \]

\[доказать.\ \]

\[\boxed{\text{636.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\frac{x - 4}{x - 5} + \frac{x - 6}{x + 5} =\]

\[= 2\ \ \ \ \ \ | \cdot (x - 5)(x + 5),\]

\[x - 5 \neq 0,\ \ x \neq 5\]

\[x + 5 \neq 0,\ \ x \neq - 5\ \]

\[(x - 4)(x + 5) +\]

\[+ (x - 6)(x - 5) = 2 \cdot \left( x^{2} - 25 \right)\]

\[x^{2} + 5x - 4x - 20 + x^{2} - 5x -\]

\[- 6x + 30 = 2x^{2} - 50\]

\[- 10x = - 60\]

\[x = 6\]

\[Ответ:x = 6.\]

\[\textbf{б)}\frac{1}{2 - x} - 1 = \frac{1}{x - 2} - \frac{6 - x}{3x^{2} - 12}\]

\[2 - x \neq 0,\ \ x \neq 2\]

\[x^{2} - 4 \neq 0,\ \ x^{2} \neq 4,\]

\[\ \ x \neq \pm 2\]

\[\frac{- 1}{x - 2} - 1 = \frac{1}{x - 2} - \frac{6 - x}{3 \cdot \left( x^{2} - 4 \right)}\]

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

\[= \frac{1}{x - 2} - \frac{6 - x}{3 \cdot \left( x^{2} - 4 \right)}\]

\[3 \cdot (1 - x)(x + 2) =\]

\[= 3 \cdot (x + 2) - (6 - x)\]

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

\[= 3x + 6 - 6 + x\]

\[- 3x^{2} - 7x + 6 = 0\ \ \ \ \ | \cdot ( - 1)\]

\[3x^{2} + 7x - 6 = 0\]

\[D = 49 + 72 = 121\]

\[x_{1,2} = \frac{- 7 \pm 11}{6}\]

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

\[Ответ:x = \left\{ - 3;\frac{2}{3} \right\}.\]

\[\textbf{в)}\frac{7y - 3}{y - y^{2}} = \frac{1}{y - 1} - \frac{5}{y(y - 1)}\]

\[y(y - 1) \neq 0,\ \ y \neq 0,\]

\[\ \ y \neq 1\]

\[\frac{7y - 3}{y(1 - y)} = \frac{1}{y - 1} - \frac{5}{y(y - 1)}\]

\[\frac{3 - 7y}{y(y - 1)} = \frac{1}{y - 1} - \frac{5}{y(y - 1)}\]

\[3 - 7y = y - 5\]

\[8y = 8\]

\[y = 1 - не\ подходит,\ так\ как\ \]

\[по\ ОДЗ\ y \neq 1.\]

\[Ответ:корней\ нет.\]

\[\textbf{г)}\frac{3}{y - 2} + \frac{7}{y + 2} = \frac{10}{y}\]

\[y \neq 0\]

\[y - 2 \neq 0,\ \ y \neq 2\]

\[y + 2 \neq 0,\ \ y \neq - 2\]

\[\frac{3y + 6 + 7y - 14}{y^{2} - 4} = \frac{10}{y}\]

\[y(10y - 8) = 10\left( y^{2} - 4 \right)\]

\[10y^{2} - 8y = 10y^{2} - 40\]

\[- 8y = - 40\]

\[y = 5\]

\[Ответ:y = 5.\]

\[\textbf{д)}\frac{x + 3}{x - 3} + \frac{x - 3}{x + 3} = 3\frac{1}{3}\]

\[x - 3 \neq 0,\ \ x \neq 3\]

\[x + 3 \neq 0,\ \ x \neq - 3\]

\[6x^{2} + 54 = 10x^{2} - 90\]

\[4x^{2} = 144\]

\[x^{2} = 36\]

\[x = \pm 6\]

\[Ответ:x = \left\{ - 6;6 \right\}.\]

\[\textbf{е)}\frac{5x + 7}{x - 2} - \frac{2x + 21}{x + 2} = 8\frac{2}{3}\]

\[x - 2 \neq 0,\ \ x \neq 2\]

\[x + 2 \neq 0,\ \ x \neq - 2\]

\[9x^{2} + 168 = 26x^{2} - 104\]

\[17x^{2} = 272\]

\[x^{2} = 16\]

\[x = \pm 4\]

\[Ответ:x = \left\{ - 4;4 \right\}\text{.\ }\]