Вопрос:

Решите уравнение: 3/(x-2)=2x+1.

Ответ:

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

\[ОДЗ:\ \ x \neq 2\]

\[\frac{3}{x - 2} - (2x + 1) = 0\]

\[\frac{3 - (x - 2)(2x + 1)}{x - 2} = 0\]

\[\frac{3 - (2x^{2} + x - 4x - 2)}{x - 2} = 0\]

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

\[\frac{- 2x^{2} + 3x + 5}{x - 2} = 0\]

\[- 2x^{2} + 3x + 5 = 0\]

\[D = b^{2} - 4ac =\]

\[= 9 - 4 \cdot ( - 2) \cdot 5 = 9 + 40 =\]

\[= 49\]

\[x_{1} = \frac{- 3 + 7}{- 4} = \frac{4}{- 4} = - 1\]

\[x_{2} = \frac{- 3 - 7}{- 4} = \frac{- 10}{- 4} = \frac{5}{2} = 2,5\]

\[Ответ:\ \ x = - 1\ \ \ \ и\ \ \ x = 2,5.\]

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