ĐKXĐ: $x\neq\{2;3\}$
a) $B=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}$
$=\dfrac{2x-9-(x+3)(x-3)+(2x+1)(x-2)}{(x-2)(x-3)}$
$=\dfrac{x^2-x-2}{(x-2)(x-3)}$
$=\dfrac{(x+1)(x-2)}{(x-2)(x-3)}$
$=\dfrac{x+1}{x-3}$
b) $B=\dfrac{1}{2}$
$↔ \dfrac{x+1}{x-3}=\dfrac{1}{2}$
$↔ 2(x+1)=x-3$
$↔ 2x+2=x-3$
$↔ x=-5$
c) $B=\dfrac{x-3}{x-3}+\dfrac{4}{x-3}$
$=1+\dfrac{4}{x-3}$
$B∈Z$ khi $(x-3)$ là ước nguyên của $4$
$→ (x-3)=±1,±2,±4$
$→ x∈\{4;5;1;7;-1\}$
