`"If " int(2cosx-sinx+lambda)/(cosx+sinx-2)dx=A " In"|cosx+sinx-2|+Bx+C, " then the value of " A+B+|lambda| " is"-.`

Explanation

To solve the integral problem given, we need to find the values of \( A \), \( B \), and \( \lambda \) from the equation:

\[ \int \frac{2\cos x - \sin x + \lambda}{\cos x + \sin x - 2} \, dx = A \ln |\cos x + \sin x - 2| + Bx + C \]

### Step 1: Set up the integral
Let:
\[ I = \int \frac{2\cos x - \sin x + \lambda}{\cos x + \sin x - 2} \, dx \]

### Step 2: Differentiate both sides
Differentiate both sides with respect to \( x \):
\[ \frac{dI}{dx} = \frac{2\cos x - \sin x + \lambda}{\cos x + \sin x - 2} \]

For the right-hand side, we differentiate:
\[ \frac{d}{dx} \left( A \ln |\cos x + \sin x - 2| + Bx + C \right) \]
Using the chain rule:
\[ = A \cdot \frac{1}{\cos x + \sin x - 2} \cdot \frac{d}{dx}(\cos x + \sin x - 2) + B \]
\[ = A \cdot \frac{1}{\cos x + \sin x - 2} \cdot (-\sin x + \cos x) + B \]

### Step 3: Set the derivatives equal
Now we equate both sides:
\[ \frac{2\cos x - \sin x + \lambda}{\cos x + \sin x - 2} = A \cdot \frac{-\sin x + \cos x}{\cos x + \sin x - 2} + B \]

### Step 4: Clear the denominator
Multiply through by \( \cos x + \sin x - 2 \):
\[ 2\cos x - \sin x + \lambda = A(-\sin x + \cos x) + B(\cos x + \sin x - 2) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side gives:
\[ 2\cos x - \sin x + \lambda = A\cos x - A\sin x + B\cos x + B\sin x - 2B \]
\[ = (A + B)\cos x + (-A + B)\sin x - 2B \]

### Step 6: Compare coefficients
Now we can compare coefficients on both sides:
1. Coefficient of \( \cos x \):
\[ 2 = A + B \quad \text{(1)} \]
2. Coefficient of \( \sin x \):
\[ -1 = -A + B \quad \text{(2)} \]
3. Constant term:
\[ \lambda = -2B \quad \text{(3)} \]

### Step 7: Solve the system of equations
From equation (1):
\[ B = 2 - A \]

Substituting \( B \) into equation (2):
\[ -1 = -A + (2 - A) \]
\[ -1 = 2 - 2A \]
\[ 2A = 3 \quad \Rightarrow \quad A = \frac{3}{2} \]

Now substituting \( A \) back into equation (1):
\[ 2 = \frac{3}{2} + B \quad \Rightarrow \quad B = 2 - \frac{3}{2} = \frac{1}{2} \]

Now substituting \( B \) into equation (3):
\[ \lambda = -2 \cdot \frac{1}{2} = -1 \]

### Step 8: Calculate \( A + B + |\lambda| \)
Now we can find \( A + B + |\lambda| \):
\[ A + B + |\lambda| = \frac{3}{2} + \frac{1}{2} + 1 = 2 + 1 = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]