ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 909

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

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

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

\[\frac{3x^{2} - 12 + 3x + 6 - 6 + x + 3x + 6}{3 \cdot \left( x^{2} - 4 \right)} = 0\]

\[\frac{3x^{2} + 7x - 6}{3 \cdot \left( x^{2} - 4 \right)} =\]

\[= 0\ \ \ \ \ \ | \cdot 3\left( x^{2} - 4 \right),\]

\[\text{\ \ }при\ x \neq \pm 2\]

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

\[D = 49 + 72 = 121\]

\[x_{1,2} = \frac{- 7 \pm \sqrt{121}}{2 \cdot 3} = \frac{- 7 \pm 11}{6}\]

\[x_{1} = \frac{4}{6} = \frac{2}{3}\]

\[x_{2} = - \frac{18}{6} = - 3\]

\[ОДЗ:\ \ x - 2 \neq 0,\]

\[x \neq 2\]

\[3x^{2} - 12 \neq 0\]

\[3 \cdot \left( x^{2} - 4 \right) \neq 0\]

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

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

\[x \neq \pm 2\]

\[Ответ:x = \frac{2}{3};\ x = - 3.\]