Key Concept: The General Term in Binomial Expansion

The general term, or $(r+1)^{\text{th}}$ term, in the binomial expansion of $(a+b)^n$ is given by the formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

For the given expansion $\left(\frac{4x}{5}-\frac{5}{2x}\right)^{2022}$: We have $n=2022$, $a=\frac{4x}{5}$, and $b=-\frac{5}{2x}$.

---

Step 1: Determine the $1011^{\text{th}}$ term from the beginning

To find the $1011^{\text{th}}$ term from the beginning, we set $r+1 = 1011$, which implies $r = 1010$. Substituting these values into the general term formula: $T_{1011} = \binom{2022}{1010} \left(\frac{4x}{5}\right)^{2022-1010} \left(-\frac{5}{2x}\right)^{1010}$ $T_{1011} = \binom{2022}{1010} \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}$

---

Step 2: Determine the $1011^{\text{th}}$ term from the end

There are two common ways to find a term from the end:

1. Counting from the beginning: The $k^{\text{th}}$ term from the end in the expansion of $(a+b)^n$ is equivalent to the $(n-k+2)^{\text{th}}$ term from the beginning. Here, $n=2022$ and $k=1011$. So, the $1011^{\text{th}}$ term from the end is the $(2022 - 1011 + 2)^{\text{th}} = 1013^{\text{th}}$ term from the beginning. For this term, $r+1 = 1013$, which means $r = 1012$. Substituting into the general term formula: $T'_{1011} = T_{1013} = \binom{2022}{1012} \left(\frac{4x}{5}\right)^{2022-1012} \left(-\frac{5}{2x}\right)^{1012}$ $T'_{1011} = \binom{2022}{1012} \left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012}$ Tip: Recall the property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$. Therefore, $\binom{2022}{1012} = \binom{2022}{2022-1012} = \binom{2022}{1010}$. This equality will be crucial for simplifying the equation.

2. Swapping $a$ and $b$: The $k^{\text{th}}$ term from the end of $(a+b)^n$ is the $k^{\text{th}}$ term from the beginning of $(b+a)^n$. This method simplifies the power distribution. However, for consistency, we'll continue with the first method as it directly relates to the $T_{r+1}$ formula with original $a$ and $b$.

---

Step 3: Set up the equation based on the given condition

The problem states that the $1011^{\text{th}}$ term from the end is 1024 times the $1011^{\text{th}}$ term from the beginning. $T'_{1011} = 1024 \times T_{1011}$ Substitute the expressions for $T'_{1011}$ and $T_{1011}$ we derived: $\binom{2022}{1012} \left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012} = 1024 \times \binom{2022}{1010} \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}$

---

Step 4: Simplify the equation

We use the binomial coefficient property $\binom{2022}{1012} = \binom{2022}{1010}$ to cancel the coefficients from both sides. $\left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012} = 1024 \times \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}$ To further simplify, we divide both sides by the common terms with the lower powers: $\left(\frac{4x}{5}\right)^{1010}$ and $\left(-\frac{5}{2x}\right)^{1010}$. $\frac{\left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012}}{\left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1010}} = 1024 \times \frac{\left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}}{\left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1010}}$ This simplifies to: $\left(-\frac{5}{2x}\right)^{1012-1010} = 1024 \times \left(\frac{4x}{5}\right)^{1012-1010}$ $\left(-\frac{5}{2x}\right)^{2} = 1024 \times \left(\frac{4x}{5}\right)^{2}$ Explanation: When squaring a negative term, the result is positive, e.g., $(-A)^2 = A^2$. $\frac{(-5)^2}{(2x)^2} = 1024 \times \frac{(4x)^2}{5^2}$ $\frac{25}{4x^2} = 1024 \times \frac{16x^2}{25}$ Common Mistake: Forgetting that $1024 = 2^{10}$. Recognizing powers of 2 (or other common bases) can simplify calculations.

---

Step 5: Solve for $|x|$

Now, we need to isolate $x^2$ and then solve for $|x|$. Multiply both sides by $4x^2 \times 25$ to clear the denominators: $25 \times 25 = 1024 \times 16x^2 \times 4x^2$ $5^2 \times 5^2 = 2^{10} \times (2^4) \times (2^2) \times x^2 \times x^2$ $5^4 = 2^{10} \times 2^{4+2} \times x^{2+2}$ $5^4 = 2^{10} \times 2^6 \times x^4$ $5^4 = 2^{16} \times x^4$ Now, isolate $x^4$: $x^4 = \frac{5^4}{2^{16}}$ To find $|x|$, we take the fourth root of both sides. $\sqrt[4]{x^4} = \sqrt[4]{\frac{5^4}{2^{16}}}$ $|x| = \frac{\sqrt[4]{5^4}}{\sqrt[4]{2^{16}}}$ $|x| = \frac{5}{2^{16/4}}$ $|x| = \frac{5}{2^4}$ $|x| = \frac{5}{16}$

---

Summary and Key Takeaway

The absolute value of $x$ is $\frac{5}{16}$. This problem effectively tests the understanding of the general term in a binomial expansion, how to find terms from the end, and algebraic simplification involving powers. The key takeaway is to carefully apply the binomial theorem, utilize properties of binomial coefficients ($\binom{n}{r} = \binom{n}{n-r}$), and meticulously simplify the resulting algebraic equation, paying attention to signs and powers.