Express the given fraction as a sum of partial fractions:

[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{A}{x + 1} + \frac{B}{3x - 5} + \frac{C}{(3x - 5)^2} \][/tex]



Answer :

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].