Докажите тождество 228

Решите задачу:

$\displaystyle \left(\frac{8y^2+2y}{8y^3-1}-\frac{2y+1}{4y^2+2y+1}\right)\cdot\left(1+ \frac{2y+1}{2y}-\frac{4y^2+10y}{4y^2+2y}\right):\frac{1}{2y}= \\ \\ \left(\frac{8y^2+2y}{(2y-1)(4y^2+2y+1)}-\frac{2y+1}{4y^2+2y+1}\right)\cdot \\ \\ \left(1+ \frac{2y+1}{2y}-\frac{4y^2+10y}{2y(2y+1)}\right):\frac{1}{2y}= \\ \\ \frac{8y^2+2y-(2y+1)(2y-1)}{(2y-1)(4y^2+2y+1)}\cdot \\ \\ \frac{2y(2y+1)+(2y+1)^2-4y^2-10y}{2y(2y+1)}: \frac{1}{2y}=$
$\displaystyle \frac{8y^2+2y-4y^2+1}{(2y-1)(4y^2+2y+1)}\cdot \frac{4y^2+2y+4y^2+4y+1-4y^2-10y}{2y(2y+1)}:\frac{1}{2y}= \\ \\ \frac{4y^2+2y+1}{(2y-1)(4y^2+2y+1)}\cdot \frac{4y^2-4y+1}{2y(2y+1)}: \frac{1}{2y}= \\ \\ \frac{1}{2y-1}\cdot \frac{(2y-1)^2}{2y(2y+1)}\cdot2y= \frac{2y-1}{2y+1}$