Question:

Step 1: Write down the general term of the binomial expansion. The general term, $T_{r+1}$, in the expansion of $(a+b)^n$ is given by the formula: $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ Step 2: Substitute the given values into the general term formula. For the given expansion $\left(\dfrac{4x}{5}+\dfrac{5}{2x^2}\right)^9$, we have $a = \dfrac{4x}{5}$, $b = \dfrac{5}{2x^2}$, and $n=9$. Substituting these values, the general term is: $$T_{r+1} = \binom{9}{r} \left(\frac{4x}{5}\right)^{9-r} \left(\frac{5}{2x^2}\right)^r$$ Step 3: Simplify the general term to separate the powers of $x$. $$T_{r+1} = \binom{9}{r} \frac{4^{9-r} x^{9-r}}{5^{9-r}} \cdot \frac{5^r}{2^r (x^2)^r}$$ $$T_{r+1} = \binom{9}{r} \frac{(2^2)^{9-r} 5^r}{5^{9-r} 2^r} \cdot \frac{x^{9-r}}{x^{2r}}$$ $$T_{r+1} = \binom{9}{r} \frac{2^{18-2r} 5^r}{5^{9-r} 2^r} \cdot x^{9-r-2r}$$ $$T_{r+1} = \binom{9}{r} 2^{18-2r-r} 5^{r-(9-r)} x^{9-3r}$$ $$T_{r+1} = \binom{9}{r} 2^{18-3r} 5^{2r-9} x^{9-3r}$$ Step 4: Determine the value of $r$ for the required power of $x$. We need to find the coefficient of $x^{-6}$. Therefore, we set the exponent of $x$ in the general term equal to $-6$: $$9-3r = -6$$ $$3r = 9 - (-6)$$ $$3r = 15$$ $$r = 5$$ Step 5: Calculate the coefficient by substituting the value of $r$. Substitute $r=5$ back into the coefficient part of the general term (excluding the $x$ term): $$\text{Coefficient} = \binom{9}{5} 2^{18-3(5)} 5^{2(5)-9}$$ $$\text{Coefficient} = \binom{9}{5} 2^{18-15} 5^{10-9}$$ $$\text{Coefficient} = \binom{9}{5} 2^3 5^1$$ First, 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 value back: $$\text{Coefficient} = 126 \times 8 \times 5$$ $$\text{Coefficient} = 126 \times 40$$ $$\text{Coefficient} = 5040$$ Step 6: State the final answer. The coefficient of $x^{-6}$ in the expansion of $\left(\dfrac{4x}{5}+\dfrac{5}{2x^2}\right)^9$ is $5040$. The final answer is $\boxed{5040}$.