ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 880

\[\boxed{\text{880.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{8x^{2} - 3}{5} - \frac{5 - 9x^{2}}{4} = 2\]

\[32x² - 12 - 25 + 45x^{2} = 2 \cdot 20\]

\[77x^{2} - 37 = 40\]

\[77x^{2} = 77\]

\[x^{2} = 1\]

\[x = \sqrt{1}\]

\[x = \pm 1\]

\[Ответ:x = \pm 1.\]

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

\[\text{\ \ }ОДЗ:x + 1 \neq 0,\ \ x \neq - 1\]

\[\frac{2x + 2 - x^{2} + x - 1}{x^{3} + 1} =\]

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

\[- x^{2} + 3x + 1 = 2x - 1\]

\[- x^{2} + 3x + 1 - 2x + 1 = 0\]

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

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

\[D = 1 + 8 = 9\]

\[x_{1,\ 2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} = - 1;2.\]

\[x_{1} = - 1 \Longrightarrow не\ подходит\]

\[\ по\ ОДЗ.\]

\[x_{2} = 2\]

\[Ответ:x = 2.\]

\[\textbf{в)}\ \frac{10}{x^{2} - 4} - \frac{3}{2x - 4} = \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\frac{10}{(x - 2)(x + 2)} - \frac{3}{2 \cdot (x - 2)} = \frac{1}{2}\]

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

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

\[20 - 3x - 6 = x^{2} - 4\]

\[x^{2} + 3x - 4 - 14 = 0\]

\[x^{2} + 3x - 18 = 0\]

\[D = 9 + 72 = 81\]

\[x_{1,2} = \frac{- 3 \pm \sqrt{81}}{2} = \frac{- 3 \pm 9}{2} = - 6;3\]

\[x_{1} = - 6;\ \ \ \ \ \ \ x_{2} = 3\ \]

\[ОДЗ:\ \]

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

\[x^{2} \neq 4\]

\[x \neq \pm 2\ \]

\[2x - 4 \neq 0\]

\[2x \neq 4\]

\[x \neq 2\]

\[Ответ:x = - 6;\ \ x = 3.\]

\[\textbf{г)}\ x - \frac{x^{2} - 17}{x - 3} = \frac{5}{x}\]

\[x\left( x^{2} - 3x - x^{2} + 17 \right) =\]

\[= 5 \cdot (x - 3)\]

\[- 3x^{2} + 17x = 5x - 15\]

\[- 3x^{2} + 12x + 15 = 0\ \ \ \ \ \ |\ :( - 3)\]

\[x^{2} - 4x - 5 = 0\]

\[D = 16 + 20 = 36\]

\[x_{1,2} = \frac{4 \pm 6}{2} = - 1;5\]

\[x_{1} = - 1,\ \ x_{2} = 5\]

\[ОДЗ:\]

\[x - 3 \neq 0\]

\[x \neq 3\]

\[x \neq 0\]

\[Ответ:x = - 1;\ \ x = 5.\]