Решите уравнение: (9x+3)/(1+3x)=x-7.

\[\frac{9x + 3}{1 + 3x} = x - 7\]

\[ОДЗ:\ \ x = - \frac{1}{3}\]

\[\frac{9x + 3}{1 + 3x} - (x - 7) = 0\]

\[\frac{9x + 3 - (x - 7)(1 + 3x)}{1 + 3x} = 0\]

\[\frac{9x + 3 - \left( x + 3x^{2} - 7 - 21x \right)}{1 + 3x} =\]

\[= 0\]

\[\frac{9x + 3 - 3x^{2} + 20x + 7}{1 + 3x} = 0\]

\[\frac{- 3x^{2} + 29x + 10}{1 + 3x} = 0\]

\[- 3x^{2} + 29x + 10 = 0\]

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

\[= 841 - 4 \cdot ( - 3) \cdot 10 =\]

\[= 841 + 120 = 961\]

\[x_{1} = \frac{- 29 + 31}{- 6} = \frac{2}{- 6} =\]

\[= - \frac{1}{3}\ (не\ подходит\ по\ ОДЗ)\]

\[x_{2} = \frac{- 29 - 31}{- 6} = \frac{- 60}{- 6} = 10\]

\[Ответ:\ \ x = 10.\]