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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 634
Задание 634
\[\boxed{\text{634.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ \frac{3x + 1}{x + 2} - \frac{x - 1}{x - 2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
\[x + 2 \neq 0\ \ \ и\ \ \ x - 2 \neq 0\]
\[\ \ \ \ \ \ \ \ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 2\]
\[3x^{2} - 6x + x - 2 - x^{2} - 2x +\]
\[+ x + 2 = x^{2} - 4\]
\[x^{2} - 6x + 4 = 0\]
\[D = 36 - 16 = 20\]
\[x_{1,2} = \frac{6 \pm \sqrt{20}}{2} =\]
\[= \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}\]
\[Ответ:x = \left\{ 3 - \sqrt{5};3 + \sqrt{5} \right\}.\]
\[\textbf{б)}\frac{2y - 2}{y + 3} + \frac{y + 3}{y - 3} = 5\]
\[y + 3 \neq 0\ \ \ и\ \ \ y - 3 \neq 0\]
\[\ \ \ \ \ \ \ \ y \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq 3\]
\[2y^{2} - 6y - 2y + 6 + y^{2} +\]
\[+ 6y + 9 = 5y^{2} - 45\]
\[- 2y^{2} - 2y + 60 = 0\ \ \ \ |\ :( - 2)\]
\[y^{2} + y - 30 = 0\]
\[D = 1 + 120 = 121\]
\[y_{1,2} = \frac{- 1 \pm 11}{2} = 5;\ - 6.\]
\[Ответ:y = \left\{ - 6;5 \right\}.\]
\[\textbf{в)}\ \frac{4}{9y^{2} - 1} - \frac{4}{3y + 1} = \frac{5}{1 - 3y}\]
\[9y^{2} - 1 \neq 0,\ \ 9y^{2} \neq 1,\]
\[\text{\ \ }y^{2} \neq \frac{1}{9},\ \ y \neq \pm \frac{1}{3}\]
\[3y + 1 \neq 0,\ \ 3y \neq - 1,\ \]
\[\ y \neq - \frac{1}{3}\]
\[1 - 3y \neq 0,\ \ - 3y \neq - 1,\]
\[\ \ y \neq \frac{1}{3}\]
\[\frac{4}{(3y - 1)(3y + 1)} -\]
\[- \frac{4}{3y + 1} = \frac{- 5}{3y - 1}\]
\[4 - 4 \cdot (3y - 1) = - 5 \cdot (3y + 1)\]
\[4 - 12y + 4 = - 15y - 5\]
\[3y = - 13\]
\[y = - \frac{13}{3}\]
\[Ответ:y = - 4\frac{1}{3}.\]
\[\textbf{г)}\frac{4}{x + 3} - \frac{5}{3 - x} = \frac{1}{x - 3} - 1\]
\[x + 3 \neq 0,\ \ x \neq - 3\]
\[x - 3 \neq 0,\ \ x \neq 3\]
\[\frac{4}{x + 3} + \frac{5}{x - 3} = \frac{1 - x + 3}{x - 3}\]
\[4 \cdot (x - 3) + 5 \cdot (x + 3) =\]
\[= (4 - x)(x + 3)\]
\[4x - 12 + 5x + 15 =\]
\[= 4x + 12 - x^{2} - 3x\]
\[x^{2} + 8x - 9 = 0\]
\[D = 64 + 36 = 100\]
\[x_{1,2} = \frac{- 8 \pm 10}{2} = 1;\ - 9.\]
\[Ответ:x = \left\{ - 9;1 \right\}.\]
\[\textbf{д)}\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x^{2} - x}\]
\[x \neq 0\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x(x - 1)}\]
\[3x - 3 + 4x = 5 - x\]
\[8x = 8\]
\[x = 1,\ так\ как\ по\ ОДЗ\ x \neq 1,\ \]
\[то\ корней\ нет.\]
\[Ответ:корней\ нет.\]
\[\textbf{е)}\frac{3y - 2}{y} - \frac{1}{y - 2} = \frac{3y + 4}{y^{2} - 2y}\]
\[y \neq 0\]
\[y - 2 \neq 0,\ \ y \neq 2\]
\[\frac{3y - 2}{y} - \frac{1}{y - 2} = \frac{3y + 4}{y(y - 2)}\]
\[(3y - 2)(y - 2) - y = 3y + 4\]
\[3y^{2} - 6y - 2y + 4 - y =\]
\[= 3y + 4\]
\[3y^{2} - 12y = 0\]
\[3y(y - 4) = 0\]
\[3y = 0\ \ \ и\ \ \ y - 4 = 0\]
\[\ \ y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = 4\]
\[y = 0 - не\ подходит\ \]
\[по\ ОДЗ\ y \neq 0\]
\[Ответ:y = 4.\ \]
