Answer
$\dfrac{\dfrac{1}{x}-\dfrac{1}{x+1}}{\dfrac{1}{x}+\dfrac{1}{x+1}}=\dfrac{1}{2x+1}$
Work Step by Step
$\dfrac{\dfrac{1}{x}-\dfrac{1}{x+1}}{\dfrac{1}{x}+\dfrac{1}{x+1}}$
Evaluate the subtraction indicated in the numerator and the sum indicated on the denominator:
$\dfrac{\dfrac{1}{x}-\dfrac{1}{x+1}}{\dfrac{1}{x}+\dfrac{1}{x+1}}=\dfrac{\dfrac{(x+1)-x}{x(x+1)}}{\dfrac{(x+1)+x}{x(x+1)}}=...$
Evaluate the division and simplify:
$...=\dfrac{(x+1)-x}{x(x+1)}\div\dfrac{(x+1)+x}{x(x+1)}=\dfrac{[(x+1)-x][x(x+1)]}{[(x+1)+x][x(x+1)]}=...$
$...=\dfrac{x+1-x}{x+1+x}=\dfrac{1}{2x+1}$