Answer
$\dfrac{(4y+3)}{(2y-1)}$
Work Step by Step
Since,in the given rational expressions denominator are same, thus take it as common and subtract the rational terms by changing the sign of each term of the second rational.
Given: $\dfrac{20y^2+5y+1}{6y^2+y-2}-\dfrac{8y^2-12y-5}{6y^2+y-2}$
or, $=\dfrac{20y^2+5y+1-8y^2+12y+5}{6y^2+y-2}$
or, $=\dfrac{12y^2+9y+8y+6}{(3y+2)(2y-1)}$
or, $=\dfrac{(3y+2)(4y+3)}{(3y+2)(2y-1)}$
or, $=\dfrac{(4y+3)}{(2y-1)}$
