Solve Complex Fraction Equation: Finding A, B, C in (A/x + Bx/2) = (2x-3)²/x - C

$\frac{A}{x}+\frac{Bx}{2}=\frac{(2x-3)^2}{x}-c$

Calculate the values of A, B, and C.

To solve this problem, we'll proceed with the following steps:

• Step 1: Expand the expression $(2x-3)^2$ on the right-hand side.
• Step 2: Simplify the resulting expression by dividing by $x$.
• Step 3: Compare the terms on both sides of the equation to deduce the coefficients $A$, $B$, and $C$.

Now, let us work through these steps:

Step 1: Expand the expression $(2x-3)^2$.

$(2x - 3)^2 = 4x^2 - 12x + 9$

Step 2: Substitute back into the equation and simplify by dividing by $x$:

$\frac{A}{x} + \frac{Bx}{2} = \frac{4x^2 - 12x + 9}{x} - C$

This simplifies to:

$4x - 12 + \frac{9}{x} - C$

Step 3: To find $A$, $B$, and $C$, compare coefficients of similar terms:

• Constant term (no $x$): Equate $-12$ with $-C$ to get $C = 12$ (note: should consider sign reversal).
• Coefficient of $x$: Compare $\frac{B}{2}$ to 4. Giving $B = 8$.
• Coefficient of $\frac{1}{x}$: Compare $A$ to 9, giving $A = 9$.

Therefore, the solution to the problem is that the values are $\textbf{A = 9, B = 8, C = -12}$.