Answer
$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}=\dfrac{x+3}{x+1}$
Work Step by Step
$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}$
Evaluate the sum present in the numerator and substraction present in the denominator:
$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}=\dfrac{\dfrac{(x+2)+1}{x+2}}{\dfrac{(x+2)-1}{x+2}}=...$
Evaluate the division:
$...=\dfrac{[(x+2)+1](x+2)}{[(x+2)-1](x+2)}=...$
Simplify:
$...=\dfrac{x+2+1}{x+2-1}=\dfrac{x+3}{x+1}$
