If `t_(r)` denotes the rth term in the expansion `(x+1/x)^(23)`, and `t_(12)=4t_(13)`, then `absx`=_____

Answer

To solve the problem, we need to find the absolute value of \( x \) given the condition \( t_{12} = 4t_{13} \) in the expansion of \( (x + \frac{1}{x})^{23} \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The \( r \)-th term in the expansion of \( (x + \frac{1}{x})^{n} \) is given by:
\[ t_r = \binom{n}{r-1} x^{n - (r-1)} \left(\frac{1}{x}\right)^{r-1} = \binom{n}{r-1} x^{n - r + 1} \cdot \frac{1}{x^{r-1}} = \binom{n}{r-1} x^{n - 2r + 2} \]
For \( n = 23 \), we have:
\[ t_r = \binom{23}{r-1} x^{23 - 2r + 2} = \binom{23}{r-1} x^{25 - 2r} \]

2. **Write the Terms \( t_{12} \) and \( t_{13} \)**:
- For \( t_{12} \):
\[ t_{12} = \binom{23}{11} x^{25 - 24} = \binom{23}{11} x^1 = \binom{23}{11} x \]
- For \( t_{13} \):
\[ t_{13} = \binom{23}{12} x^{25 - 26} = \binom{23}{12} x^{-1} = \binom{23}{12} \frac{1}{x} \]

3. **Set Up the Equation**:
According to the problem, we have:
\[ t_{12} = 4t_{13} \]
Substituting the expressions for \( t_{12} \) and \( t_{13} \):
\[ \binom{23}{11} x = 4 \cdot \binom{23}{12} \cdot \frac{1}{x} \]

4. **Cross-Multiply**:
\[ \binom{23}{11} x^2 = 4 \cdot \binom{23}{12} \]

5. **Use the Relationship Between Binomial Coefficients**:
We know that:
\[ \binom{23}{12} = \frac{23 - 12 + 1}{12} \cdot \binom{23}{11} = \frac{12}{12} \cdot \binom{23}{11} = \binom{23}{11} \]
Thus, we can simplify:
\[ \binom{23}{11} x^2 = 4 \cdot \binom{23}{11} \]

6. **Divide Both Sides by \( \binom{23}{11} \)** (assuming it is not zero):
\[ x^2 = 4 \]

7. **Solve for \( x \)**:
\[ x = \pm 2 \]

8. **Find the Absolute Value**:
\[ |x| = 2 \]

### Final Answer:
Thus, the absolute value of \( x \) is:
\[ \boxed{2} \]