a) Ta tìm $m$ thỏa mãn
$\dfrac{3m-5}{m-1} + \dfrac{2m-5}{2-m} = 1$
$<-> \dfrac{(3m-5)(2-m) + (2m-5)(m-1)}{(m-1)(2-m)} =1$
$<-> (-3m^2 +11m - 10) + (2m^2 -7m + 5) = -m^2 +3m - 2$
$<-> m = 3$
Vậy $m = 3$
b) Ta tìm $m$ thỏa mãn
$\dfrac{9m-7}{3m-2} + \dfrac{5-4m}{2m-3} = 1$
$<-> \dfrac{9m-7}{3m-2} - \dfrac{4m-5}{2m-3} = 1$
$<-> \dfrac{(9m-7)(2m-3) - (4m-5)(3m-2)}{(3m-2)(2m-3)} = 1$
$<-> 18m^2 -41m + 21 - (12m^2-23m + 10) = 6m^2 -13m + 6$
$<-> -5m = -5$
$<-> m = 1$
Vậy $m = 1$.