$B=\dfrac{2a-27}{a-12}=\dfrac{4}{5}$
$\dfrac{5(2a-27)}{5(a-12)}$, $\dfrac{4(a-12)}{5(a-12)}$
$\Rightarrow 5(2a-27)=4(a-12)$
$\Rightarrow 10a-135=4a-48$
$\Rightarrow 6a=87$
$\Rightarrow a=\dfrac{87}{6}=\dfrac{29}{2}$
b) Đặt $t=\dfrac{2a-27}{a-12}$
$\Rightarrow \dfrac{1}{t}=\dfrac{1}{\dfrac{2a-27}{a-12}}=\dfrac{a-12}{2a-27}$
$\Rightarrow t=\dfrac{1}{t}$
$\Rightarrow t^2=1\Rightarrow t=\pm1$
TH1: $t=1\Rightarrow \dfrac{2a-27}{a-12}=1$
$\Rightarrow 2a-27=1\Rightarrow x=\dfrac{27+1}{2}=14$
TH2: $t=-1\Rightarrow \dfrac{2a-27}{a-12}=-1$
$\Rightarrow 2a-27=-1(a-12)$
$\Rightarrow 2a-27=12-a$
$\Rightarrow 3a=39\Rightarrow a=13$.
c) $B=\dfrac{2a-17}{a-12}=\dfrac{2a-24-3}{a-12}=2-\dfrac{3}{a-12}$
Để $B$ thuộc $\mathbb Z$ thì $3 $ $\vdots$ $ (a-12)$
$\Rightarrow a-12=Ư(3)=\{\pm1;\pm3\}$
Với $a-12=1\Rightarrow a=13$ (tm)
Với $a-12=-1\Rightarrow a=11$ (tm)
Với $a-12=3\Rightarrow a=15$ (tm)
Với $a-12=-3\Rightarrow a=9$ (tm)
d) GTLN $B=5$ khi $\dfrac{3}{a-12}=-3\Rightarrow a=9$
