Answer
$\dfrac{1+\dfrac{1}{y-2}}{y+\dfrac{1}{y-2}}=\dfrac{1}{y-1}$
Work Step by Step
$\dfrac{1+\dfrac{1}{y-2}}{y+\dfrac{1}{y-2}}$
Evaluate the sums indicated in the numerator and in the denominator:
$\dfrac{1+\dfrac{1}{y-2}}{y+\dfrac{1}{y-2}}=\dfrac{\dfrac{y-2+1}{y-2}}{\dfrac{y(y-2)+1}{y-2}}=\dfrac{\dfrac{y-1}{y-2}}{\dfrac{y^{2}-2y+1}{y-2}}=...$
Evaluate the division:
$...=\dfrac{y-1}{y-2}\div\dfrac{y^{2}-2y+1}{y-2}=\dfrac{(y-1)(y-2)}{(y^{2}-2y+1)(y-2)}=...$
$...=\dfrac{y-1}{y^{2}-2y+1}=...$
Factor the denominator, which is a perfect square trinomial, and simplify:
$...=\dfrac{y-1}{(y-1)^{2}}=\dfrac{1}{y-1}$