Найдите корни уравнения: 3/(1-4y^2)+4/(2y^2+y)=3/(4y^2+4y+1).

Ответ:

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

\[= \frac{3}{4y² + 4y + 1}\]

\[ОДЗ:\ \ y \neq \pm \frac{1}{2}\]

\[\ \ \ \ \ \ \ \ \ \ \ \ y \neq 0\]

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

\[= \frac{3}{(2y + 1)²}\]

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

\[(3y + 4 - 8y)(2y + 1) =\]

\[= 3y(1 - 2y)\]

\[( - 5y + 4)(2y + 1) = 3y - 6y^{2}\]

\[- 10y^{2} - 5y + 8y + 4 - 3y + 6y^{2} = 0\]

\[- 4y^{2} + 4 = 0\]

\[- 4 \cdot \left( y^{2} - 1 \right) = 0\ \]

\[y^{2} - 1 = 0\]

\[y^{2} = 1\]

\[y = \pm 1\]

\[Ответ:y = \pm 1.\]