Question

мне бы это уравнение решить даю десятку

Answer 1

Ответ:

$\frac{20y}{36 {y}^{2} - 4} - \frac{2y - 3}{2 - 6y} = \frac{5 - 2y}{6y + 2} \\ \frac{5y}{9 {y}^{2} - 1} - \frac{2y - 3}{2 - 6y} - \frac{5 - 2y}{6y + 2} = 0 \\ \frac{5y}{(3y - 1)(3y + 1)} + \frac{2y - 3}{2(3y - 1)} - \frac{5 - 2y}{2(3y + 1)} = 0 \\ \frac{10y + (3y + 1)(2y - 3) - (3y - 1)(5 - 2y)}{2(3y - 1)(3y + 1)} = 0 \\ \frac{10y + 6 {y}^{2} - 9y + 2y - 3 - 17y + 6 {y}^{2} + 5}{2(3y - 1)(3y + 1)} = 0 \\ \frac{ - 14y + 12 {y}^{2} + 2 }{2(3y - 1)(3y + 1)} = 0 \\ \frac{2( - 7y + 6 {y}^{2} + 1) }{2(3y - 1)(3y + 1)} = 0 \\ \frac{ - 7y + 6y {}^{2} + 1}{(3y - 1)(3y + 1)} = 0 \\ - 7y + 6 {y}^{2} + 1 = 0 \\ 6y {}^{2} - 7y + 1 = 0 \\ 6 {y }^{2} - y - 6y + 1 = 0 \\ y(6y - 1) - (6y - 1) = 0 \\ (6y - 1)(y - 1) = 0 \\ \\ 6y - 1 = 0 \\ y - 1 = 0 \\ y1 = \frac{1}{6} \\ y2 = 1$