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To solve the equation
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{A}{x + 1} + \frac{B}{3x - 5} + \frac{C}{(3x - 5)^2} \][/tex]
we need to decompose the left side into a sum of partial fractions. Follow these steps:
### Step 1: Set Up the Partial Fractions
Firstly, you suppose that:
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{A}{x + 1} + \frac{B}{3x - 5} + \frac{C}{(3x - 5)^2} \][/tex]
### Step 2: Clear the Denominators
Multiply both sides by the common denominator [tex]\((x + 1)(3x - 5)^2\)[/tex] to clear the denominators:
[tex]\[ 2x - 1 = A(3x - 5)^2 + B(x + 1)(3x - 5) + C(x + 1) \][/tex]
### Step 3: Expand and Combine Like Terms
Expand the right-hand side:
1. Expand [tex]\(A(3x - 5)^2\)[/tex]:
[tex]\[ A(3x - 5)^2 = A(9x^2 - 30x + 25) = 9Ax^2 - 30Ax + 25A \][/tex]
2. Expand [tex]\(B(x + 1)(3x - 5)\)[/tex]:
[tex]\[ B(x + 1)(3x - 5) = B(3x^2 - 5x + 3x - 5) = B(3x^2 - 2x - 5) = 3Bx^2 - 2Bx - 5B \][/tex]
3. Expand [tex]\(C(x + 1)\)[/tex]:
[tex]\[ C(x + 1) = Cx + C \][/tex]
So, combining all these terms:
[tex]\[ 2x - 1 = (9A + 3B)x^2 + (-30A - 2B + C)x + (25A - 5B + C) \][/tex]
### Step 4: Equate Coefficients
Now, equate the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant terms on both sides:
#### Coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ 0 = 9A + 3B \][/tex]
#### Coefficient of [tex]\(x\)[/tex]:
[tex]\[ 2 = -30A - 2B + C \][/tex]
#### Constant term:
[tex]\[ -1 = 25A - 5B + C \][/tex]
### Step 5: Solve the System of Equations
Solve the system of linear equations:
1. [tex]\(9A + 3B = 0\)[/tex]: This simplifies to [tex]\(3A + B = 0\)[/tex]
2. [tex]\( -30A - 2B + C = 2\)[/tex]
3. [tex]\(25A - 5B + C = -1\)[/tex]
First, from [tex]\(3A + B = 0\)[/tex]:
[tex]\[ B = -3A \][/tex]
Substitute [tex]\(B = -3A\)[/tex] into the second and third equations:
For [tex]\( -30A - 2B + C = 2\)[/tex]:
[tex]\[ -30A - 2(-3A) + C = 2\\ -30A + 6A + C = 2\\ -24A + C = 2\\ C = 2 + 24A \][/tex]
For [tex]\( 25A - 5B + C = -1\)[/tex]:
[tex]\[ 25A - 5(-3A) + C = -1\\ 25A + 15A + C = -1\\ 40A + C = -1\\ C = -1 - 40A \][/tex]
Now, equate the two expressions for [tex]\(C\)[/tex]:
[tex]\[ 2 + 24A = -1 - 40A \][/tex]
Solve for [tex]\(A\)[/tex]:
[tex]\[ 2 + 24A = -1 - 40A\\ 64A = -3\\ A = -\frac{3}{64} \][/tex]
Since [tex]\(B = -3A\)[/tex]:
[tex]\[ B = -3 \left( -\frac{3}{64} \right) = \frac{9}{64} \][/tex]
And for [tex]\(C\)[/tex]:
[tex]\[ C = 2 + 24 \left( -\frac{3}{64} \right)\\ C = 2 - \frac{72}{64}\\ C = 2 - \frac{9}{8}\\ C = \frac{7}{8} \][/tex]
### Conclusion
The coefficients are:
[tex]\[ A = -\frac{3}{64}, \quad B = \frac{9}{64}, \quad C = \frac{7}{8} \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{-\frac{3}{64}}{x + 1} + \frac{\frac{9}{64}}{3x - 5} + \frac{\frac{7}{8}}{(3x - 5)^2} \][/tex]
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{A}{x + 1} + \frac{B}{3x - 5} + \frac{C}{(3x - 5)^2} \][/tex]
we need to decompose the left side into a sum of partial fractions. Follow these steps:
### Step 1: Set Up the Partial Fractions
Firstly, you suppose that:
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{A}{x + 1} + \frac{B}{3x - 5} + \frac{C}{(3x - 5)^2} \][/tex]
### Step 2: Clear the Denominators
Multiply both sides by the common denominator [tex]\((x + 1)(3x - 5)^2\)[/tex] to clear the denominators:
[tex]\[ 2x - 1 = A(3x - 5)^2 + B(x + 1)(3x - 5) + C(x + 1) \][/tex]
### Step 3: Expand and Combine Like Terms
Expand the right-hand side:
1. Expand [tex]\(A(3x - 5)^2\)[/tex]:
[tex]\[ A(3x - 5)^2 = A(9x^2 - 30x + 25) = 9Ax^2 - 30Ax + 25A \][/tex]
2. Expand [tex]\(B(x + 1)(3x - 5)\)[/tex]:
[tex]\[ B(x + 1)(3x - 5) = B(3x^2 - 5x + 3x - 5) = B(3x^2 - 2x - 5) = 3Bx^2 - 2Bx - 5B \][/tex]
3. Expand [tex]\(C(x + 1)\)[/tex]:
[tex]\[ C(x + 1) = Cx + C \][/tex]
So, combining all these terms:
[tex]\[ 2x - 1 = (9A + 3B)x^2 + (-30A - 2B + C)x + (25A - 5B + C) \][/tex]
### Step 4: Equate Coefficients
Now, equate the coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant terms on both sides:
#### Coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ 0 = 9A + 3B \][/tex]
#### Coefficient of [tex]\(x\)[/tex]:
[tex]\[ 2 = -30A - 2B + C \][/tex]
#### Constant term:
[tex]\[ -1 = 25A - 5B + C \][/tex]
### Step 5: Solve the System of Equations
Solve the system of linear equations:
1. [tex]\(9A + 3B = 0\)[/tex]: This simplifies to [tex]\(3A + B = 0\)[/tex]
2. [tex]\( -30A - 2B + C = 2\)[/tex]
3. [tex]\(25A - 5B + C = -1\)[/tex]
First, from [tex]\(3A + B = 0\)[/tex]:
[tex]\[ B = -3A \][/tex]
Substitute [tex]\(B = -3A\)[/tex] into the second and third equations:
For [tex]\( -30A - 2B + C = 2\)[/tex]:
[tex]\[ -30A - 2(-3A) + C = 2\\ -30A + 6A + C = 2\\ -24A + C = 2\\ C = 2 + 24A \][/tex]
For [tex]\( 25A - 5B + C = -1\)[/tex]:
[tex]\[ 25A - 5(-3A) + C = -1\\ 25A + 15A + C = -1\\ 40A + C = -1\\ C = -1 - 40A \][/tex]
Now, equate the two expressions for [tex]\(C\)[/tex]:
[tex]\[ 2 + 24A = -1 - 40A \][/tex]
Solve for [tex]\(A\)[/tex]:
[tex]\[ 2 + 24A = -1 - 40A\\ 64A = -3\\ A = -\frac{3}{64} \][/tex]
Since [tex]\(B = -3A\)[/tex]:
[tex]\[ B = -3 \left( -\frac{3}{64} \right) = \frac{9}{64} \][/tex]
And for [tex]\(C\)[/tex]:
[tex]\[ C = 2 + 24 \left( -\frac{3}{64} \right)\\ C = 2 - \frac{72}{64}\\ C = 2 - \frac{9}{8}\\ C = \frac{7}{8} \][/tex]
### Conclusion
The coefficients are:
[tex]\[ A = -\frac{3}{64}, \quad B = \frac{9}{64}, \quad C = \frac{7}{8} \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2x - 1}{(x + 1)(3x - 5)^2} = \frac{-\frac{3}{64}}{x + 1} + \frac{\frac{9}{64}}{3x - 5} + \frac{\frac{7}{8}}{(3x - 5)^2} \][/tex]
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