Answer
$(f+g)(x)=\dfrac{x^2+2x-1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(f-g)(x)=\dfrac{x^2+1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(f\cdot g)(x)=\dfrac{x+1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(\dfrac{f}{g})(x)=\dfrac{x^2+x}{x-1}\space\space\{x|x\ne 1 \}$
Work Step by Step
$(f+g)(x)=\dfrac{x+1}{x-1}+\dfrac{1}{x}$
$=\dfrac{x+1}{x-1}\cdot\dfrac{x}{x}+\dfrac{1}{x}\cdot\dfrac{x-1}{x-1}$
$=\dfrac{x^2+x}{x^2-x}+\dfrac{x-1}{x^2-x}$
$=\dfrac{x^2+2x-1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(f-g)(x)=\dfrac{x+1}{x-1}-\dfrac{1}{x}$
$=\dfrac{x+1}{x-1}\cdot\dfrac{x}{x}-\dfrac{1}{x}\cdot\dfrac{x-1}{x-1}$
$=\dfrac{x^2+x}{x^2-x}-\dfrac{x-1}{x^2-x}$
$=\dfrac{x^2+1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(f\cdot g)(x)=\dfrac{x+1}{x-1}\cdot\dfrac{1}{x}$
$=\dfrac{x+1}{x^2-x}\space\space\{x|x\ne0, 1 \}$
$(\dfrac{f}{g})(x)=\dfrac{x+1}{x-1}\div\dfrac{1}{x}$
$=\dfrac{x^2+x}{x-1}\space\space\{x|x\ne 1 \}$
