To expand \((2 + 3x)^{-2}\), we can use the binomial series expansion for negative exponents:

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

Let \(u = \frac{3}{2}x\) and \(n = 2\), so \((2 + 3x)^{-2} = \left(2(1 + \frac{3}{2}x)\right)^{-2} = \frac{1}{4}(1 + \frac{3}{2}x)^{-2}\).

Now expand \((1 + \frac{3}{2}x)^{-2}\):

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

\(= 1 - 3x + \frac{9}{2}x^2\)

Multiply by \(\frac{1}{4}\):

\(\frac{1}{4}(1 - 3x + \frac{9}{2}x^2) = \frac{1}{4} - \frac{3}{4}x + \frac{27}{16}x^2\)

(i) We start by simplifying \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})\).

This is a difference of squares: \((a+b)(a-b) = a^2 - b^2\).

Thus, \((\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x\).

Therefore, \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x}) = 2x\).

Now, deduce that:

\(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} = \frac{\sqrt{1+x} - \sqrt{1-x}}{2x}.\)

(ii) To find the expansion of \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}}\), we use the binomial expansion for \(\sqrt{1+x}\) and \(\sqrt{1-x}\).

\(\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\)

\(\sqrt{1-x} \approx 1 - \frac{x}{2} - \frac{x^2}{8}\)

Adding these gives:

\(\sqrt{1+x} + \sqrt{1-x} \approx 2 - \frac{x^2}{4}\)

Thus, \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} \approx \frac{1}{2 - \frac{x^2}{4}}\).

Using the binomial expansion for \(\frac{1}{1-y}\), where \(y = \frac{x^2}{8}\), we get:

\(\frac{1}{2} \left(1 + \frac{x^2}{8}\right)^{-1} \approx \frac{1}{2} \left(1 - \frac{x^2}{8}\right)\)

\(= \frac{1}{2} - \frac{1}{16}x^2\)

(a) We start by rewriting \((2 - x^2)^{-2}\) as \(\left(\frac{1}{2}(1 - \frac{x^2}{2})\right)^{-2}\).

Using the binomial expansion for \((1 + u)^n\), where \(n = -2\) and \(u = -\frac{x^2}{2}\), we have:

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

Substitute \(n = -2\) and \(u = -\frac{x^2}{2}\):

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

\(= 1 + x^2 + \frac{3}{4}x^4 + \cdots\)

Multiply by \(\frac{1}{4}\) to account for the factor \(\frac{1}{4}\) from \(\left(\frac{1}{2}\right)^{-2}\):

\(\frac{1}{4}(1 + x^2 + \frac{3}{4}x^4) = \frac{1}{4} + \frac{1}{4}x^2 + \frac{3}{16}x^4\)

(b) The expansion is valid for \(|x| < \sqrt{2}\) because the binomial expansion is valid for \(|u| < 1\), where \(u = \frac{x^2}{2}\), leading to \(|x^2/2| < 1\), or \(|x| < \sqrt{2}\).

To expand \((1 + 4x)^{-\frac{1}{2}}\), we use the binomial series expansion for \((1 + u)^n\), where \(u = 4x\) and \(n = -\frac{1}{2}\).

The binomial series expansion is:

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

Substitute \(n = -\frac{1}{2}\) and \(u = 4x\):

\(1 + \left(-\frac{1}{2}\right)(4x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(4x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(4x)^3\)

Calculate each term:

First term: \(1\)

Second term: \(-2x\)

Third term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(16x^2) = 6x^2\)

Fourth term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{6}(64x^3) = -20x^3\)

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

\(1 - 2x + 6x^2 - 20x^3\)

To expand \(\frac{1}{(2+x)^3}\), we can rewrite it as \((2+x)^{-3}\) and use the binomial series expansion for \((1+u)^n\), where \(u = \frac{x}{2}\) and \(n = -3\).

The binomial expansion is given by:

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

Substituting \(u = \frac{x}{2}\) and \(n = -3\), we have:

\((1+\frac{x}{2})^{-3} = 1 - 3\left(\frac{x}{2}\right) + \frac{-3(-4)}{2}\left(\frac{x}{2}\right)^2 + \cdots\)

Simplifying each term:

First term: \(1\)

Second term: \(-3 \times \frac{x}{2} = -\frac{3}{2}x\)

Third term: \(\frac{12}{2} \times \frac{x^2}{4} = 3 \times \frac{x^2}{4} = \frac{3}{4}x^2\)

Thus, the expansion of \((1+\frac{x}{2})^{-3}\) up to \(x^2\) is:

\(1 - \frac{3}{2}x + \frac{3}{4}x^2\)

Now, multiply by \(\frac{1}{8}\) to account for the factor of \(\frac{1}{8}\) in the original expression:

\(\frac{1}{8}(1 - \frac{3}{2}x + \frac{3}{4}x^2) = \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2\)