We are asked to express the rational function $\frac{4x-9}{(x-2)(x-3)}$ in partial fractions.

Okay, let's solve these partial fraction decomposition problems. Due to the extensive nature of the problems, I will select problem (a) as a representative example to demonstrate the solution process.

1. Problem Description

We are asked to express the rational function $\frac{4x-9}{(x-2)(x-3)}$ in partial fractions.

2. Solution Steps

We want to find constants $A$ and $B$ such that:

$\frac{4x-9}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$

To find $A$ and $B$, we multiply both sides by $(x-2)(x-3)$:

$4x-9 = A(x-3) + B(x-2)$

We can solve for $A$ and $B$ by substituting suitable values for $x$.

If $x=2$:

$4(2) - 9 = A(2-3) + B(2-2)$

$8 - 9 = -A$

$-1 = -A$

$A = 1$

If $x=3$:

$4(3) - 9 = A(3-3) + B(3-2)$

$12 - 9 = B$

$3 = B$

$B = 3$

So, we have $A = 1$ and $B = 3$. Therefore, the partial fraction decomposition is:

$\frac{4x-9}{(x-2)(x-3)} = \frac{1}{x-2} + \frac{3}{x-3}$

3. Final Answer

$\frac{4x-9}{(x-2)(x-3)} = \frac{1}{x-2} + \frac{3}{x-3}$