Решите систему неравенств: 6x^2-7x+1<0; 4x-3<0.

Ответ:

\[\left\{ \begin{matrix} 6x^{2} - 7x + 1 < 0 \\ 4x - 3 < \ 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[6x^{2} - 7x + 1 = 6\left( x + \frac{1}{6} \right)(x - 1)\]

\[D = 49 - 24 = 25\]

\[x_{1} = \frac{7 + 5}{12} = 1;\]

\[x_{2} = \frac{7 - 5}{12} = - \frac{2}{12} = - \frac{1}{6};\]

\[\left\{ \begin{matrix} \left( x + \frac{1}{6} \right)(x - 1) < 0 \\ 4x < 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - \frac{1}{6} < x < 1 \\ x < \frac{3}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[- \frac{1}{6} < x < \frac{3}{4}.\]