Let `f^(1)(x)=(3x+2)/(2x+3),x in R-{(-3)/(2)}`
For `nge2,` define `f^(n(x))=f^(1)of^(n-1)(x)`.
If `f^(5)(x)=(ax+b)/(bx+a),gcd(a,b)=1`,then `a+b` is equal to

Answer

To solve the problem, we need to find \( f^5(x) \) based on the function \( f^1(x) = \frac{3x + 2}{2x + 3} \) and then express it in the form \( \frac{ax + b}{bx + a} \) where \( \gcd(a, b) = 1 \). Finally, we will compute \( a + b \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function given:
\[ f^1(x) = \frac{3x + 2}{2x + 3} \]

2. **Calculate \( f^2(x) \)**:
To find \( f^2(x) \), we need to compute \( f^1(f^1(x)) \):
\[ f^2(x) = f^1(f^1(x)) = f^1\left(\frac{3x + 2}{2x + 3}\right) \]
Substitute \( x \) in \( f^1(x) \):
\[ f^1\left(\frac{3x + 2}{2x + 3}\right) = \frac{3\left(\frac{3x + 2}{2x + 3}\right) + 2}{2\left(\frac{3x + 2}{2x + 3}\right) + 3} \]
Simplifying the numerator:
\[ = \frac{\frac{9x + 6 + 4x + 6}{2x + 3}}{\frac{6x + 4 + 6x + 9}{2x + 3}} = \frac{(9x + 4x + 12)}{(6x + 6x + 13)} = \frac{13x + 12}{12x + 13} \]

3. **Calculate \( f^3(x) \)**:
Now, we find \( f^3(x) = f^1(f^2(x)) \):
\[ f^3(x) = f^1\left(\frac{13x + 12}{12x + 13}\right) \]
Substitute into \( f^1 \):
\[ = \frac{3\left(\frac{13x + 12}{12x + 13}\right) + 2}{2\left(\frac{13x + 12}{12x + 13}\right) + 3} \]
Simplifying:
\[ = \frac{\frac{39x + 36 + 24x + 26}{12x + 13}}{\frac{26x + 24 + 36x + 39}{12x + 13}} = \frac{(39x + 24x + 62)}{(26x + 36x + 63)} = \frac{63x + 62}{62x + 63} \]

4. **Calculate \( f^4(x) \)**:
Continuing, we find \( f^4(x) = f^1(f^3(x)) \):
\[ f^4(x) = f^1\left(\frac{63x + 62}{62x + 63}\right) \]
Substitute:
\[ = \frac{3\left(\frac{63x + 62}{62x + 63}\right) + 2}{2\left(\frac{63x + 62}{62x + 63}\right) + 3} \]
Simplifying gives:
\[ = \frac{(189x + 186 + 124)}{(126x + 124 + 189)} = \frac{189x + 186 + 124}{126x + 189 + 124} = \frac{313x + 310}{315x + 313} \]

5. **Calculate \( f^5(x) \)**:
Finally, we find \( f^5(x) = f^1(f^4(x)) \):
\[ f^5(x) = f^1\left(\frac{313x + 310}{315x + 313}\right) \]
Substitute:
\[ = \frac{3\left(\frac{313x + 310}{315x + 313}\right) + 2}{2\left(\frac{313x + 310}{315x + 313}\right) + 3} \]
Simplifying gives:
\[ = \frac{(939x + 930 + 620)}{(626x + 620 + 939)} = \frac{939x + 930 + 620}{626x + 939 + 620} = \frac{1559x + 1550}{1565x + 1559} \]

6. **Identify \( a \) and \( b \)**:
From \( f^5(x) = \frac{1559x + 1550}{1565x + 1559} \), we can identify:
- \( a = 1559 \)
- \( b = 1550 \)

7. **Calculate \( a + b \)**:
Now, we compute:
\[ a + b = 1559 + 1550 = 3109 \]

### Final Answer:
Thus, the value of \( a + b \) is \( \boxed{3109} \).