## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}=\dfrac{x+3}{x+1}$
$\dfrac{1+\dfrac{1}{x+2}}{1-\dfrac{1}{x+2}}$ 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: $...=\dfrac{x+2+1}{x+2-1}=\dfrac{x+3}{x+1}$