To solve the equation [tex]\(\frac{2 x - 1}{(x + 1)(3 x - 5)^2} = \frac{A}{(x + 1)} + \frac{B}{(3 x - 5)} + \frac{C}{(3 x - 5)^2}\)[/tex], we need to perform partial fraction decomposition. Let's go through the steps methodically.
### Step 1: Establish a common denominator
The given equation already provides the common denominator [tex]\((x + 1)(3 x - 5)^2\)[/tex].
### Step 2: Express the right-hand side with the common denominator
Write the given equation with a single common denominator:
[tex]\[
\frac{A}{(x + 1)} + \frac{B}{(3 x - 5)} + \frac{C}{(3 x - 5)^2} = \frac{A (3 x - 5)^2 + B (x + 1) (3 x - 5) + C (x + 1)}{(x + 1) (3 x - 5)^2}
\][/tex]
### Step 3: Equate the numerators
Set the numerators of both sides equal to each other:
[tex]\[
2 x - 1 = A (3 x - 5)^2 + B (x + 1) (3 x - 5) + C (x + 1)
\][/tex]
### Step 4: Expand and simplify the right side
First, expand [tex]\(A (3 x - 5)^2\)[/tex]:
[tex]\[
A (3 x - 5)^2 = A (9 x^2 - 30 x + 25)
\][/tex]
Next, expand [tex]\(B (x + 1) (3 x - 5)\)[/tex]:
[tex]\[
B (x + 1) (3 x - 5) = B (3 x^2 - 2 x - 5)
\][/tex]
Finally, expand [tex]\(C (x + 1)\)[/tex]:
[tex]\[
C (x + 1) = C x + C
\][/tex]
Putting these together, we get:
[tex]\[
A (9 x^2 - 30 x + 25) + B (3 x^2 - 2 x - 5) + C x + C
\][/tex]
### Step 5: Combine like terms
Combine all the polynomial terms from the right-hand side:
[tex]\[
(9 A + 3 B)x^2 + (-30 A - 2 B + C)x + (25 A - 5 B + C)
\][/tex]
This should equal the left side [tex]\(2 x - 1\)[/tex]:
[tex]\[
(9 A + 3 B)x^2 + (-30 A - 2 B + C)x + (25 A - 5 B + C) = 2 x - 1
\][/tex]
### Step 6: Match coefficients of polynomial terms
For the coefficients to match, we need:
- Coefficient of [tex]\(x^2\)[/tex]: [tex]\(9A + 3B = 0\)[/tex]
- Coefficient of [tex]\(x\)[/tex]: [tex]\(-30A - 2B + C = 2\)[/tex]
- Constant term: [tex]\(25A - 5B + C = -1\)[/tex]
### Step 7: Solve the system of linear equations
Solve these equations for [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[
\begin{cases}
9A + 3B = 0 \\
-30A - 2B + C = 2 \\
25A - 5B + C = -1
\end{cases}
\][/tex]
Upon solving this system, we find:
[tex]\[
\begin{cases}
A = 0 \\
B = 0 \\
C = 0
\end{cases}
\][/tex]
### Conclusion
Thus, the coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] all turn out to be zero. Therefore, there are no solutions for [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] that satisfy the given equation. Therefore, the solution is [tex]\([]\)[/tex].
