Let `D_1=|a b a+b c d c+d a b a-b|a n dD_2=|a c a+c b d b+d a c a+b+c|` then the value of `|(D_1)/(D_2)|,w h e r eb!=0a n da d!=b c ,` is _____.

Explanation

To solve the given problem, we need to evaluate the determinants \( D_1 \) and \( D_2 \) and then find the value of \( \left| \frac{D_1}{D_2} \right| \).

### Step 1: Define the determinants

We have:
\[ D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix} \]
and
\[ D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix} \]

### Step 2: Simplify \( D_1 \)

We will apply a column operation on \( D_1 \). Specifically, we will replace the third column with the result of the operation \( C_3 - C_1 + C_2 \):
\[ C_3 \rightarrow C_3 - C_1 + C_2 \]

This gives us:
\[ D_1 = \begin{vmatrix} a & b & 0 \\ c & d & 0 \\ a & b & -2b \end{vmatrix} \]

### Step 3: Expand \( D_1 \)

Now, we can expand \( D_1 \) along the third column:
\[ D_1 = -2b \begin{vmatrix} a & b \\ c & d \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]
Thus, we have:
\[ D_1 = -2b(ad - bc) \]

### Step 4: Simplify \( D_2 \)

Next, we will apply the same column operation on \( D_2 \):
\[ C_3 \rightarrow C_3 - C_1 + C_2 \]
This gives us:
\[ D_2 = \begin{vmatrix} a & c & 0 \\ b & d & 0 \\ a & c & b \end{vmatrix} \]

### Step 5: Expand \( D_2 \)

Now, we can expand \( D_2 \) along the third column:
\[ D_2 = b \begin{vmatrix} a & c \\ b & d \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} a & c \\ b & d \end{vmatrix} = ad - bc \]
Thus, we have:
\[ D_2 = b(ad - bc) \]

### Step 6: Calculate \( \left| \frac{D_1}{D_2} \right| \)

Now, we can find \( \left| \frac{D_1}{D_2} \right| \):
\[ \frac{D_1}{D_2} = \frac{-2b(ad - bc)}{b(ad - bc)} = -2 \]
Taking the modulus:
\[ \left| \frac{D_1}{D_2} \right| = |-2| = 2 \]

### Final Answer

Thus, the value of \( \left| \frac{D_1}{D_2} \right| \) is \( \boxed{2} \).