Answer
$\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{3m-2}{m(m-1)}$
or
$\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{3m-2}{m^{2}-m}$
Work Step by Step
$\dfrac{1}{m-1}+\dfrac{2}{m}$
Evaluate the sum of these two fractions by using the LCD, which is $m(m-1)$ in this case:
$\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{m+2(m-1)}{m(m-1)}=...$
Simplify:
$...=\dfrac{m+2m-2}{m(m-1)}=\dfrac{3m-2}{m(m-1)}$
