The function \(f\left( x \right) = \frac{1}{{x - 1}}\) we should have \(x - 1 \ne 0 \Leftrightarrow x \ne 1\).
The function \(g\left( x \right) = \frac{1}{x} - 2\) , we should have \(x \ne 0\).
a)
Obtain the function \(\left( {f + g} \right)\left( x \right)\):
\(\begin{aligned}\left( {f + g} \right)\left( x \right) &= \frac{1}{{x - 1}} + \frac{1}{x} - 2\\ &= \frac{{x + x - 1 - 2x\left( {x - 1} \right)}}{{x\left( {x - 1} \right)}}\\ &= \frac{{2x - 1 - 2{x^2} + 2x}}{{{x^2} - x}}\\ &= - \frac{{2{x^2} - 4x + 1}}{{{x^2} - x}},\left\{ {x|x \ne 0,1} \right\}\end{aligned}\)
b)
Obtain the function \(\left( {f - g} \right)\left( x \right)\):
\(\begin{aligned}\left( {f - g} \right)\left( x \right) &= \frac{1}{{x - 1}} - \frac{1}{x} - 2\\ &= \frac{{x - \left( {x - 1} \right) + 2x\left( {x - 1} \right)}}{{x\left( {x - 1} \right)}}\\ &= \frac{{1 + 2{x^2} - 2x}}{{{x^2} - x}}\\ &= \frac{{2{x^2} - 2x + 1}}{{{x^2} - x}},\left\{ {x|x \ne 0,1} \right\}\end{aligned}\)
c)
Obtain the function \(\left( {fg} \right)\left( x \right)\):
\(\begin{aligned}\left( {fg} \right)\left( x \right) &= \frac{1}{{x - 1}}\left( {\frac{1}{x} - 2} \right)\\ &= \frac{1}{{{x^2} - x}} - \frac{2}{{x - 1}}\\ &= \frac{{1 - 2x}}{{{x^2} - x}},\left\{ {x|x \ne 0,1} \right\}\end{aligned}\)
d)
Obtain the function \(\left( {\frac{f}{g}} \right)\left( x \right)\):
\(\begin{aligned}\left( {\frac{f}{g}} \right)\left( x \right) &= \frac{{\frac{1}{{x - 1}}}}{{\frac{1}{x} - 2}}\\ &= \frac{1}{{x - 1}} \cdot \frac{x}{{1 - 2x}}\\ &= \frac{x}{{\left( {x - 1} \right)\left( {1 - 2x} \right)}}\\ &= - \frac{x}{{\left( {x - 1} \right)\left( {2x - 1} \right)}}\\ &= - \frac{x}{{2{x^2} - 3x + 1}},\left\{ {x|x \ne 0,\frac{1}{2},1} \right\}\end{aligned}\)
The additional domain restriction is \(g\left( x \right) \ne 0 \Rightarrow x \ne \frac{1}{2}\).