Решите уравнение (3y+2)/(4y^2+y)-(3-y)/(16y^2-1)+3/(1-4y)=0.

Ответ:

\[\frac{3y + 2}{4y^{2} + y} - \frac{3 - y}{16y^{2} - 1} + \frac{3}{1 - 4y} = 0\]

\[\frac{3y + 2^{\backslash 4y - 1}}{y(4y + 1)} - \frac{3 - y^{\backslash y}}{(4y - 1)(4y + 1)} - \frac{3^{\backslash 4y^{2} + y}}{4y - 1} = 0\]

\[\frac{12y^{2} + 8y - 3y - 2 - 3y + y^{2} - 12y^{2} - 3y}{y(4y + 1)(4y - 1)} = 0\]

\[ОДЗ:\ \ y \neq 0;\ \ y \neq \pm \frac{1}{4}.\]

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

\[y_{1} + y_{2} = 1;\ \ \ y_{1} \cdot y_{2} = - 2\]

\[y_{1} = 2;\ \ y_{2} = - 1.\]

\[Ответ:y_{1} = 2;\ \ y_{2} = - 1.\]