If `(7x-1)/(6x^(2)-5x+1)=(A)/(3x-1)+(B)/(2x-1)`, then `A+B=?`.

Answer

To solve the equation \(\frac{7x-1}{6x^2-5x+1} = \frac{A}{3x-1} + \frac{B}{2x-1}\), we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \(6x^2 - 5x + 1\). We can rewrite it as:
\[ 6x^2 - 5x + 1 = (3x - 1)(2x - 1) \]

### Step 2: Rewrite the Equation
Now, we can rewrite the equation using the factored form:
\[ \frac{7x - 1}{(3x - 1)(2x - 1)} = \frac{A}{3x - 1} + \frac{B}{2x - 1} \]

### Step 3: Combine the Right Side
To combine the right side, we will find a common denominator:
\[ \frac{A(2x - 1) + B(3x - 1)}{(3x - 1)(2x - 1)} \]

### Step 4: Set the Numerators Equal
Since the denominators are the same, we can set the numerators equal to each other:
\[ 7x - 1 = A(2x - 1) + B(3x - 1) \]

### Step 5: Expand the Right Side
Expanding the right side gives:
\[ 7x - 1 = (2Ax - A) + (3Bx - B) = (2A + 3B)x - (A + B) \]

### Step 6: Compare Coefficients
Now we can compare the coefficients of \(x\) and the constant terms:
1. For \(x\): \(2A + 3B = 7\)
2. For the constant term: \(-A - B = -1\) or \(A + B = 1\)

### Step 7: Solve the System of Equations
We now have a system of equations:
1. \(2A + 3B = 7\)
2. \(A + B = 1\)

From the second equation, we can express \(A\) in terms of \(B\):
\[ A = 1 - B \]

Substituting this into the first equation:
\[ 2(1 - B) + 3B = 7 \]
\[ 2 - 2B + 3B = 7 \]
\[ 2 + B = 7 \]
\[ B = 5 \]

Now substituting \(B = 5\) back into \(A + B = 1\):
\[ A + 5 = 1 \implies A = 1 - 5 = -4 \]

### Step 8: Find \(A + B\)
Now we can find \(A + B\):
\[ A + B = -4 + 5 = 1 \]

### Final Answer
Thus, the value of \(A + B\) is:
\[ \boxed{1} \]