We start by using the binomial expansion for \((1 + u)^n\), where \(u = -2x\) and \(n = -\frac{1}{2}\).

The binomial expansion is given by:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substituting \(u = -2x\) and \(n = -\frac{1}{2}\), we have:

\((1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(-2x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(-2x)^3 + \cdots\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{1}{2} \times 2x = x\)

Third term: \(\frac{3}{8} \times 4x^2 = \frac{3}{2}x^2\)

Fourth term: \(\frac{15}{48} \times 8x^3 = \frac{5}{2}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 + x + \frac{3}{2}x^2 + \frac{5}{2}x^3\)