सोडवा
`(5y^(2)+40y-12)/(5y+10y^(2)-4) =(y+8)/(1+2y)`

Solve: (5y^2+40y-12)/(5y+10y^2-4)=(y+8)/(1+2y)

Divide: (2y^(2) - 3y^(3) + 5y-4) // (y-1)

The centres of the circles x ^(2) + y ^(2) =1, x ^(2) + y^(2)+ 6x - 2y =1 and x ^(2) + y^(2) -12 x + 4y =1 are

Simplify : (y^2-3y+2)/(y^2+y-2)div(2y^2-5y+2)/(2y^2+3y-2)