Expand the first three terms of (1+(x)/(2))10 using the binomial theorem.

The binomial theorem states that $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots$.
Here, $a=1$, $b=\frac{x}{2}$, and $n=10$.

The first term is:

$$\binom{10}{0}(1)^{10}\left(\frac{x}{2}\right)^0 = 1 \cdot 1 \cdot 1 = 1$$

The second term is:

$$\binom{10}{1}(1)^{9}\left(\frac{x}{2}\right)^1 = 10 \cdot 1 \cdot \frac{x}{2} = 5x$$

The third term is:

$$\binom{10}{2}(1)^{8}\left(\frac{x}{2}\right)^2 = \frac{10 \times 9}{2 \times 1} \cdot 1 \cdot \frac{x^2}{4} = 45 \cdot \frac{x^2}{4} = \frac{45}{4}x^2$$

So, the expansion of $\left(1+\frac{x}{2}\right)^{10}$ up to the third term is $1 + 5x + \frac{45}{4}x^2$.

We need to find the first three terms of $(1-2x)\left(1 + 5x + \frac{45}{4}x^2 + \dots\right)$.
Multiply each term in $(1-2x)$ by the terms in the expansion, keeping only terms up to $x^2$.

Multiply 1 by the expansion:

$$1 \cdot \left(1 + 5x + \frac{45}{4}x^2\right) = 1 + 5x + \frac{45}{4}x^2$$

Multiply $-2x$ by the expansion:

$$-2x \cdot \left(1 + 5x + \frac{45}{4}x^2\right) = -2x - 10x^2 - \frac{45}{2}x^3$$

We only need terms up to $x^2$, so we ignore $-\frac{45}{2}x^3$.

Add the results from Step 2:

$$(1 + 5x + \frac{45}{4}x^2) + (-2x - 10x^2)$$

Combine like terms:

$$1 + (5x - 2x) + \left(\frac{45}{4}x^2 - 10x^2\right)$$ $$1 + 3x + \left(\frac{45}{4} - \frac{40}{4}\right)x^2$$ $$1 + 3x + \frac{5}{4}x^2$$

The first three terms in the binomial expansion are 1, $3x$, and $\frac{5}{4}x^2$.