Question:

Step 1: Write down the general term of the binomial expansion. The given binomial expansion is $\left(\dfrac{x^{5/2}}{2}-\dfrac{4}{x^\ell}\right)^9$. Using the binomial theorem, the general term $T_{r+1}$ in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here, $a = \dfrac{x^{5/2}}{2}$, $b = -\dfrac{4}{x^\ell}$, and $n=9$. Substituting these values, we get: $$T_{r+1} = \binom{9}{r} \left(\dfrac{x^{5/2}}{2}\right)^{9-r} \left(-\dfrac{4}{x^\ell}\right)^r$$ Step 2: Simplify the general term to separate powers of $x$ and constant factors. We separate the coefficients and the powers of $x$: $$T_{r+1} = \binom{9}{r} \frac{(x^{5/2})^{9-r}}{2^{9-r}} (-4)^r \frac{1}{(x^\ell)^r}$$ $$T_{r+1} = \binom{9}{r} \frac{x^{\frac{5}{2}(9-r)}}{2^{9-r}} (-1)^r (2^2)^r x^{-\ell r}$$ $$T_{r+1} = \binom{9}{r} (-1)^r \frac{2^{2r}}{2^{9-r}} x^{\frac{45}{2}-\frac{5r}{2}-\ell r}$$ $$T_{r+1} = \binom{9}{r} (-1)^r 2^{2r-(9-r)} x^{\frac{45}{2}-\left(\frac{5}{2}+\ell\right)r}$$ $$T_{r+1} = \binom{9}{r} (-1)^r 2^{3r-9} x^{\frac{45}{2}-\left(\frac{5}{2}+\ell\right)r}$$ Step 3: Use the condition for the constant term to find the value of $\ell$. For the term to be constant, the exponent of $x$ must be zero: $$\frac{45}{2}-\left(\frac{5}{2}+\ell\right)r = 0$$ Multiplying by 2, we get: $$45 - (5+2\ell)r = 0 \quad (*)$$ The constant term is given as $-84$. So, the coefficient part must be $-84$: $$\binom{9}{r} (-1)^r 2^{3r-9} = -84$$ We look for an integer value of $r$ (from $0$ to $9$) that satisfies this. Let's try $r=3$. If $r=3$, the coefficient becomes: $$\binom{9}{3} (-1)^3 2^{3(3)-9} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times (-1) \times 2^{9-9} = 84 \times (-1) \times 2^0 = -84$$ This matches the given constant term. So, $r=3$ for the constant term. Now, substitute $r=3$ into equation $(*)$: $$45 - (5+2\ell)(3) = 0$$ $$45 = 3(5+2\ell)$$ $$15 = 5+2\ell$$ $$10 = 2\ell$$ $$\ell = 5$$ Step 4: Determine the exponent of $x$ for which the coefficient is required. We need to find the coefficient of $x^{-3\ell}$. Since we found $\ell=5$, we need the coefficient of $x^{-3(5)} = x^{-15}$. The exponent of $x$ in the general term is $\frac{45}{2}-\left(\frac{5}{2}+\ell\right)r$. Substitute $\ell=5$: $$\frac{45}{2}-\left(\frac{5}{2}+5\right)r = \frac{45}{2}-\left(\frac{5}{2}+\frac{10}{2}\right)r = \frac{45}{2}-\frac{15}{2}r$$ Set this exponent equal to $-15$: $$\frac{45}{2}-\frac{15}{2}r = -15$$ $$45 - 15r = -30$$ $$15r = 75$$ $$r = 5$$ Step 5: Calculate the coefficient of $x^{-3\ell}$. For $r=5$ (and $\ell=5$), the coefficient is given by $\binom{9}{r} (-1)^r 2^{3r-9}$: $$\binom{9}{5} (-1)^5 2^{3(5)-9}$$ Calculate $\binom{9}{5}$: $$\binom{9}{5} = \binom{9}{9-5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$$ Now substitute this back into the coefficient expression: $$126 \times (-1) \times 2^{15-9} = 126 \times (-1) \times 2^6$$ $$= -126 \times 64$$ To calculate $126 \times 64$: $126 \times 64 = 8064$ So, the coefficient of $x^{-3\ell}$ is $-8064$. Step 6: Express the coefficient in the form $2^\alpha\beta$ and determine $\alpha$ and $\beta$. The problem states the coefficient is $2^\alpha\beta$, where $\beta<0$ is an odd number. We have $-8064 = 2^\alpha\beta$. Let's find the prime factorization of $8064$: $8064 = 2 \times 4032$ $4032 = 2 \times 2016$ $2016 = 2 \times 1008$ $1008 = 2 \times 504$ $504 = 2 \times 252$ $252 = 2 \times 126$ $126 = 2 \times 63$ So, $8064 = 2^7 \times 63$. Therefore, $-8064 = -2^7 \times 63$. Comparing this with $2^\alpha\beta$, we have: $\alpha = 7$ $\beta = -63$ This satisfies the conditions that $\beta < 0$ and $\beta$ is an odd number. Step 7: Calculate the final value of $|\alpha\ell-\beta|$. We have $\alpha=7$, $\ell=5$, and $\beta=-63$. Substitute these values into the expression $|\alpha\ell-\beta|$: $$|\alpha\ell-\beta| = |(7)(5) - (-63)|$$ $$= |35 + 63|$$ $$= |98|$$ $$= 98$$ The final answer is $\boxed{98}$.