The sum of the reciprocals of the roots of the equation `abx^2=(a^2+b^2+2ab) (x-1)` is:

Options

A. 2
B. `1/2`
C. `2/3`
D. 1

Answer

To find the sum of the reciprocals of the roots of the equation \( abx^2 = (a^2 + b^2 + 2ab)(x - 1) \), we can follow these steps:

### Step 1: Rearrange the equation
First, we need to rearrange the given equation into a standard quadratic form \( Ax^2 + Bx + C = 0 \).

Starting with:
\[ abx^2 = (a^2 + b^2 + 2ab)(x - 1) \]

Expanding the right side:
\[ abx^2 = (a^2 + b^2 + 2ab)x - (a^2 + b^2 + 2ab) \]

Rearranging gives:
\[ abx^2 - (a^2 + b^2 + 2ab)x + (a^2 + b^2 + 2ab) = 0 \]

### Step 2: Identify coefficients
From the standard form \( Ax^2 + Bx + C = 0 \), we identify:
- \( A = ab \)
- \( B = -(a^2 + b^2 + 2ab) \)
- \( C = a^2 + b^2 + 2ab \)

### Step 3: Use the formula for the sum of the reciprocals of the roots
The sum of the reciprocals of the roots \( \alpha \) and \( \beta \) can be calculated using the formula:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \]

### Step 4: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{B}{A} = -\frac{-(a^2 + b^2 + 2ab)}{ab} = \frac{a^2 + b^2 + 2ab}{ab} \)
- The product of the roots \( \alpha \beta = \frac{C}{A} = \frac{a^2 + b^2 + 2ab}{ab} \)

### Step 5: Substitute into the formula
Now substituting into the formula for the sum of the reciprocals:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{\frac{a^2 + b^2 + 2ab}{ab}}{\frac{a^2 + b^2 + 2ab}{ab}} = 1 \]

### Final Answer
Thus, the sum of the reciprocals of the roots is:
\[ \boxed{1} \]