`(r_(2)+r_(3))sqrt((r r_(1))/(r_(2)r_(3)))=`

Options

A. a
B. b
C. c
D. bc

Explanation

To solve the equation \((r_2 + r_3) \sqrt{\frac{r r_1}{r_2 r_3}} = ?\), we will use some properties of triangles and the relationships between the inradius and exradii.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( r \) be the inradius of the triangle.
- Let \( r_1, r_2, r_3 \) be the exradii opposite to vertices A, B, and C respectively.

2. **Use the Known Formula**:
- We know that \( r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \)
- \( r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} \)
- \( r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \)
- Here, \( R \) is the circumradius of the triangle.

3. **Substituting the Values**:
- Substitute the values of \( r_1, r_2, r_3 \) into the expression:
\[ (r_2 + r_3) = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} + 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \]
- Factor out \( 4R \cos \frac{A}{2} \):
\[ = 4R \cos \frac{A}{2} \left( \sin \frac{B}{2} \cos \frac{C}{2} + \cos \frac{B}{2} \sin \frac{C}{2} \right) \]
- Using the sine addition formula, this simplifies to:
\[ = 4R \cos \frac{A}{2} \sin \left( \frac{B}{2} + \frac{C}{2} \right) \]

4. **Calculating the Square Root**:
- Now, we need to calculate \( \sqrt{\frac{r r_1}{r_2 r_3}} \):
\[ = \sqrt{\frac{r \cdot 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{(4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2})(4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2})}} \]
- Simplifying this gives:
\[ = \sqrt{\frac{r \cdot \sin \frac{A}{2}}{\cos^2 \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}} \]

5. **Final Expression**:
- Combine the results from steps 3 and 4:
\[ (r_2 + r_3) \sqrt{\frac{r r_1}{r_2 r_3}} = 4R \cos \frac{A}{2} \sin \left( \frac{B}{2} + \frac{C}{2} \right) \cdot \sqrt{\frac{r \cdot \sin \frac{A}{2}}{\cos^2 \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}} \]

6. **Using the Sine Rule**:
- Finally, we can relate this back to the sides of the triangle using the sine rule, leading us to conclude that the expression simplifies to \( a \).

### Conclusion:
Thus, the final result is \( a \).