ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 798

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\[\textbf{а)}\ \frac{x + 1}{6} + \frac{20}{x - 1} = 4\ \ \ \ \]

\[| \cdot 6(x - 1)\]

\[x - 1 \neq 0;\ \ x \neq 1\]

\[(x + 1)(x - 1) + 6 \cdot 20 =\]

\[= 24 \cdot (x - 1)\]

\[x^{2} - 1 + 120 = 24x - 24\]

\[x^{2} - 24x + 143 = 0\]

\[D_{1} = 144 - 143 = 1\]

\[x_{1} = 12 + 1 = 13;\ \ \]

\[x_{2} = 12 - 1 = 11.\]

\[Ответ:x = \left\{ 11;13 \right\}.\]

\[\textbf{б)}\ \frac{x + 15}{4} - \frac{21}{x + 2} =\]

\[= 2\ \ \ \ \ \ \ \ | \cdot 4(x + 2)\]

\[x \neq 0\]

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

\[(x + 15)(x + 2) - 21 \cdot 4 =\]

\[= 8 \cdot (x + 2)\]

\[x^{2} + 2x + 15x + 30 - 84 =\]

\[= 8x + 16\]

\[x^{2} + 9x - 70 = 0\]

\[D = 81 + 280 = 361 = 19^{2}\]

\[x_{1,2} = \frac{- 9 \pm 19}{2}\]

\[x_{1} = \frac{- 9 - 19}{2} = - 14;\ \ \ \ \]

\[x_{2} = \frac{- 9 + 19}{2} = 5\]

\[Ответ:x = - 14;\ \ x = 5.\]

\[\textbf{в)}\frac{12}{x - 1} - \frac{18}{x + 1} =\]

\[= 1\ \ \ \ \ \ | \cdot (x - 1)(x + 1)\]

\[x - 1 \neq 0,\ \ x \neq 1\]

\[x + 1 \neq 0,\ \ x \neq - 1\ \]

\[12 \cdot (x + 1) - 18 \cdot (x - 1) =\]

\[= x^{2} - 1\]

\[12x + 12 - 18x + 18 = x^{2} - 1\]

\[x^{2} - 4x - 21 = 0\]

\[D = 16 + 84 = 100\]

\[x_{1,2} = \frac{4 \pm 10}{2} = 7; - 3\]

\[Ответ:x = \left\{ - 3;7 \right\}.\]

\[\textbf{г)}\frac{16}{x - 3} + \frac{30}{1 - x} =\]

\[= 3\ \ \ \ \ \ | \cdot (x - 3)(1 - x)\]

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

\[1 - x \neq 0,\ \ x \neq 1\]

\[16 \cdot (1 - x) + 30 \cdot (x - 3) =\]

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

\[16 - 16x + 30x - 90 =\]

\[= 12x - 9 - 3x^{2}\]

\[3x^{2} + 2x - 65 = 0\]

\[D = 4 + 780 = 784 = 28^{2}\]

\[x_{1,2} = \frac{- 2 \pm 28}{6}\]

\[x_{1} = \frac{- 2 - 28}{6} = - 5;\ \ \ \]

\[x_{2} = \frac{- 2 + 28}{6} = \frac{26}{6} = 4\frac{1}{3}\]

\[Ответ:x = - 5;\ \ x = 4\frac{1}{3}.\]

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

\[= \frac{28}{1 - x^{2}}\ \ \ \ \ \ | \cdot (1 - x)^{2}\]

\[x^{2} \neq 1,\ \ x \neq \pm 1\]

\[3 \cdot (1 + x) + (1 - x) = 28\]

\[3 + 3x + 1 - x = 28\]

\[2x = 24\]

\[x = 12\]

\[Ответ:x = 12.\]

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

\[= \frac{20}{x^{2} - 4}\ \ \ \ \ \ \ | \cdot \left( x^{2} - 4 \right)\]

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

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

\[5 \cdot (x + 2) - 3 \cdot (x - 2) = 20\]

\[5x + 10 - 3x + 6 = 20\]

\[2x = 4\]

\[x = 2 - не\ подходит\ по\ ОДЗ.\]

\[Ответ:корней\ нет.\]

\[\textbf{ж)}\frac{x + 2}{x + 1} + \frac{x + 3}{x - 2} =\]

\[= \frac{29}{(x + 1)(x - 2)}\text{\ \ \ \ \ \ }\]

\[| \cdot (x + 1)(x - 2)\]

\[x + 1 \neq 0,\ \ x \neq - 1\]

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

\[(x + 2)(x - 2) +\]

\[+ (x + 3)(x + 1) = 29\]

\[x^{2} - 4 + x^{2} + x + 3x + 3 = 29\]

\[2x^{2} + 4x - 30 = 0\ \ \ \ \ \ \ |\ :2\]

\[x^{2} + 2x - 15 = 0\]

\[D = 4 + 60 = 64\]

\[x_{1,2} = \frac{- 2 \pm 8}{2} = 3;\ - 5\]

\[Ответ:x = \left\{ - 5;3 \right\}.\]

\[\textbf{з)}\frac{x + 2}{x + 3} - \frac{x + 1}{x - 1} =\]

\[= \frac{4}{(x + 3)(x - 1)}\text{\ \ \ \ \ \ }\]

\[\ \ | \cdot (x + 3)(x - 1)\]

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

\[x - 1 \neq 0,\ \ x \neq 1\]

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

\[- (x + 1)(x + 3) = 4\]

\[x^{2} - x + 2x - 2 - x^{2} -\]

\[- 3x - x - 3 = 4\]

\[- 3x = 9\]

\[x = - 3 - не\ подходит\ по\ ОДЗ\]

\[Ответ:корней\ нет.\ \]