Answer
$f+g= \frac{x^2+2x-1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
$f-g= \frac{x^2+1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
$f\cdot g= \frac{x+1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
$\frac{f}{g} =\frac{x(x+1)}{x-1}$, domain $\{x|x\ne0, 1 \}$
Work Step by Step
Step 1. Given $f(x)=\frac{x+1}{x-1}, g(x)=\frac{1}{x}$, we have $f+g=(\frac{x+1}{x-1})+(\frac{1}{x})=\frac{x^2+2x-1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
Step 2. We have $f-g=(\frac{x+1}{x-1})-(\frac{1}{x})=\frac{x^2+1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
Step 3. We have $f\cdot g=(\frac{x+1}{x-1})(\frac{1}{x})=\frac{x+1}{x(x-1)}$, domain $\{x|x\ne0, 1 \}$
Step 4. We have $\frac{f}{g}=\frac{\frac{x+1}{x-1}}{\frac{1}{x}}=\frac{x(x+1)}{x-1}$, domain $\{x|x\ne0, 1 \}$
