$\text { In, }\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}\right)^9$

$T_{ r +1}={ }^9 C_{ r } \frac{\left(x^{5 / 2}\right)^{9- r }}{2^{9- r }}\left(\frac{-4}{x^{\ell}}\right)^{ r }$

$=(-1)^{ r } \frac{{ }^9 C_{ r }}{2^{9- I }} 4^{ r } x ^{\frac{45}{2}-\frac{ r }{2} Ir }$

$=45-5 r -21 r =0$

$r =\frac{45}{5+21}……….(1)$

Now, according to the question, $(-1)^{ T } \frac{{ }^9 C _{ r }}{2^{9- I }} 4^{ T }=-84$

$=(-1)^{ I }{ }^9 C _{ r } 2^{3 T -9}=21 \times 4$

Only natural value of $r$ possible if $3 r-9=0$ $r =3$ and ${ }^9 C _3=84$

$\therefore 1=5$ from equation $(1)$

Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5 r }{2}}$-ir at $l=5$, gives

$r=5$

$\therefore{ }^9 c_5(-1) \frac{4^5}{2^4}=2^\alpha \times \beta$

$=-63 \times 2^7$

$\Rightarrow \alpha=7, \beta=-63$

$\therefore \quad \text { value of }|\alpha \ell-\beta|=98$