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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 690
Задание 690
\[\boxed{\text{690\ (690).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x + 1}{6} + \frac{20}{x - 1} = 4\ \ \ \ | \cdot 6(x - 1)\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[(x + 1)(x - 1) + 6 \cdot 20 =\]
\[= 24 \cdot (x - 1)\]
\[x^{2} - 1 + 120 = 24x - 24\]
\[x^{2} - 24x + 143 = 0\]
\[D = 576 - 572 = 4\]
\[x_{1,2} = \frac{24 \pm 2}{2} = 13;11.\]
\[Ответ:x = \left\{ 11;13 \right\}.\]
\[x \neq 0\]
\[x + 2 \neq 0,\ \ x \neq - 2\ \]
\[(x + 15)(x + 2) - 21 \cdot 4 =\]
\[= 8 \cdot (x + 2)\]
\[x^{2} + 2x + 15x + 30 - 84 =\]
\[= 8x + 16\]
\[x^{2} + 9x - 70 = 0\]
\[D = 81 + 280 = 361 = 19^{2}\]
\[x_{1,2} = \frac{- 9 \pm 19}{2}\]
\[x_{1} = \frac{- 9 - 19}{2} = - 14;\ \ \]
\[\text{\ \ }x_{2} = \frac{- 9 + 19}{2} = 5\]
\[Ответ:x = - 14;\ \ x = 5.\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[x + 1 \neq 0,\ \ x \neq - 1\ \]
\[12 \cdot (x + 1) - 18 \cdot (x - 1) =\]
\[= x^{2} - 1\]
\[12x + 12 - 18x + 18 = x^{2} - 1\]
\[x^{2} - 4x - 21 = 0\]
\[D = 16 + 84 = 100\]
\[x_{1,2} = \frac{4 \pm 10}{2} = 7; - 3\]
\[Ответ:x = \left\{ - 3;7 \right\}.\]
\[x - 3 \neq 0,\ \ x \neq 3\]
\[1 - x \neq 0,\ \ x \neq 1\]
\[16 \cdot (1 - x) + 30 \cdot (x - 3) =\]
\[= 3 \cdot (x - 3)(1 - x)\]
\[16 - 16x + 30x - 90 =\]
\[= 12x - 9 - 3x^{2}\]
\[3x^{2} + 2x - 65 = 0\]
\[D = 4 + 780 = 784 = 28^{2}\]
\[x_{1,2} = \frac{- 2 \pm 28}{6}\]
\[x_{1} = \frac{- 2 - 28}{6} = - 5;\]
\[\text{\ \ \ }x_{2} = \frac{- 2 + 28}{6} = \frac{26}{6} = 4\frac{1}{3}\]
\[Ответ:x = - 5;\ \ x = 4\frac{1}{3}.\]
\[x^{2} \neq 1,\ \ x \neq \pm 1\]
\[3 \cdot (1 + x) + (1 - x) = 28\]
\[3 + 3x + 1 - x = 28\]
\[2x = 24\]
\[x = 12\]
\[Ответ:x = 12.\]
\[x^{2} - 4 \neq 0,\ \ x^{2} \neq 4,\]
\[\ \ x \neq \pm 2\]
\[5 \cdot (x + 2) - 3 \cdot (x - 2) = 20\]
\[5x + 10 - 3x + 6 = 20\]
\[2x = 4\]
\[x = 2 - не\ подходит\ по\ ОДЗ.\]
\[Ответ:корней\ нет.\]
\[x + 1 \neq 0,\ \ x \neq - 1\]
\[x - 2 \neq 0,\ \ x \neq 2\]
\[(x + 2)(x - 2) + (x + 3)(x + 1) = 29\]
\[x^{2} - 4 + x^{2} + x + 3x + 3 = 29\]
\[2x^{2} + 4x - 30 = 0\ \ \ \ \ \ \ |\ :2\]
\[x^{2} + 2x - 15 = 0\]
\[D = 4 + 60 = 64\]
\[x_{1,2} = \frac{- 2 \pm 8}{2} = 3;\ - 5\]
\[Ответ:x = \left\{ - 5;3 \right\}.\]
\[x + 3 \neq 0,\ \ x \neq - 3\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[(x + 2)(x - 1) - (x + 1)(x + 3) = 4\]
\[- 3x = 9\]
\[x = - 3 - не\ подходит\ по\ ОДЗ\]
\[Ответ:корней\ нет.\ \]
\[\boxed{\text{690.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[x + 2y = 5\]
\[2y = - x + 5\]
\[y = - \frac{1}{2}x + \frac{5}{2}.\]
\[1)\ x + y = 5\]
\[y = - x + 5.\]
\[2)\ \frac{1}{4}y - 4x = 0\]
\[\frac{1}{4}y = 4x\]
\[y = 16x.\]
\[3)\ 6y + 3x = 10\]
\[6y = - 3x + 10\]
\[y = - \frac{1}{2}x + \frac{5}{3}.\]
\[4)\ 0,6x - 3 = - 1,2\]
\[0,6x = 1,8\]
\[x = 3.\]
\[5)\ 2x + 4y = 10\]
\[4y = - 2x + 10\]
\[y = - \frac{1}{2}x + \frac{5}{2}.\]
\[6)\ 2x + 4y = 9\]
\[4y = - 2x + 9\]
\[y = - \frac{1}{2}x + \frac{9}{4}.\]
\[7)\ 15 - 3x = 6y\]
\[y = - \frac{1}{2}x + \frac{5}{2}.\]
\[8)\ 0,5y + 0,25x = 4,8\]
\[0,5y = - 0,25x + 4,8\]
\[y = - \frac{1}{2}x + \frac{48}{5}.\]
\[Единственное\ решение\ с\ \]
\[прямыми\ 1);2);4).\]
\[Не\ имеет\ решений\ с\ \]
\[прямыми\ 3);6);8).\]
\[Бесконечно\ много\ решений\]
\[\ с\ прямыми\ 5);7).\]
