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Let's break down the problem step by step.

### Part (a)
The given expression is:
\[
(1 - 2x)^{\frac{3}{5}} = 1 + ax + bx^2 + \dots
\]
You are asked to find the values of \(a\) and \(b\).

This is essentially a binomial expansion problem where we need to expand \((1 - 2x)^{\frac{3}{5}}\) using the generalized binomial theorem for fractional powers. The binomial theorem for any power \(n\) can be written as:

\[
(1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \dots
\]

In this case, \(u = -2x\) and \(n = \frac{3}{5}\).

So, the expansion starts as:
\[
(1 - 2x)^{\frac{3}{5}} = 1 + \frac{3}{5}(-2x) + \frac{\frac{3}{5}\left(\frac{3}{5} - 1\right)}{2!}(-2x)^2 + \dots
\]

Simplifying the first two terms:

- The first term is just 1.
- The second term is:
  \[
  \frac{3}{5}(-2x) = -\frac{6}{5}x
  \]

For the third term (coefficient of \(x^2\)):

- The general term for \(x^2\) is:
  \[
  \frac{\frac{3}{5}\left(\frac{3}{5} - 1\right)}{2!}(-2x)^2 = \frac{\frac{3}{5}\left(-\frac{2}{5}\right)}{2} \cdot 4x^2 = \frac{-6}{25} \cdot 2x^2 = -\frac{12}{25}x^2
  \]

Thus, the expansion up to \(x^2\) is:
\[
(1 - 2x)^{\frac{3}{5}} = 1 - \frac{6}{5}x - \frac{12}{25}x^2 + \dots
\]

From this, we can compare it with \(1 + ax + bx^2 + \dots\), giving:
- \(a = -\frac{6}{5}\)
- \(b = -\frac{12}{25}\)

### Part (b)
Now, you are asked to use the values of \(a\) and \(b\) to find \(30^{\frac{3}{5}}\) in the form of \(\frac{m}{n}\).

We can use the binomial approximation for \(30^{\frac{3}{5}}\). Notice that:
\[
30 = 32 - 2
\]
So, we write:
\[
30^{\frac{3}{5}} = (32 - 2)^{\frac{3}{5}} = 32^{\frac{3}{5}} \left(1 - \frac{2}{32}\right)^{\frac{3}{5}}
\]

We can approximate the binomial expansion as:
\[
(1 - \frac{1}{16})^{\frac{3}{5}} \approx 1 + \left(\frac{3}{5}\right) \left(-\frac{1}{16}\right) + \dots = 1 - \frac{3}{80}
\]

Thus, we have:
\[
30^{\frac{3}{5}} \approx 32^{\frac{3}{5}} \left(1 - \frac{3}{80}\right)
\]

Next, calculate \(32^{\frac{3}{5}}\):
\[
32^{\frac{3}{5}} = 2^3 = 8
\]

Therefore:
\[
30^{\frac{3}{5}} \approx 8 \times \left(1 - \frac{3}{80}\right) = 8 \times \frac{77}{80} = \frac{616}{80} = \frac{77}{10}
\]

So, \(m = 77\) and \(n = 10\).

### Final Answers:
1. \(a = -\frac{6}{5}\), \(b = -\frac{12}{25}\)
2. \(30^{\frac{3}{5}} = \frac{77}{10}\)

Would you like a more detailed explanation on any step, or is there another part you'd like to explore further?

#### Related Questions:
1. What is the binomial expansion for \((1 + x)^n\) for any real number \(n\)?
2. How do you find higher-order terms in a binomial expansion with fractional powers?
3. How can you approximate roots using binomial expansions?
4. What happens to the accuracy of approximations as more terms are added in a binomial expansion?
5. Can the binomial theorem be applied to other functions beyond polynomials?

#### Tip:
When expanding binomials with fractional exponents, always check the radius of convergence to ensure the approximation is valid for the values of \(x\) used.

Given that (1 - 2x)^(3/5) = 1 + ax + bx^2 + ..., determine the values of a and b. Then, using the values of a and b, find the value of 30^(3/5) in the form of m/n, where m and n are integers.

Learn how to determine the values of a and b from the binomial expansion of (1 - 2x)^(3/5) using the generalized binomial theorem for fractional powers. Additionally, find the value of 30^(3/5) expressed as a fraction. This problem covers key algebraic concepts and is suitable for students in Grades 11-12.

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