Find the middle term in the expansion of `((x)/(y)+(y)/(x))^(10)`.

Answer

To find the middle term in the expansion of \(\left(\frac{x}{y} + \frac{y}{x}\right)^{10}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = \frac{x}{y}\), \(b = \frac{y}{x}\), and \(n = 10\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{10}{r} \left(\frac{x}{y}\right)^{10-r} \left(\frac{y}{x}\right)^r \]

### Step 2: Simplify the General Term
We can simplify the general term:
\[ T_{r+1} = \binom{10}{r} \left(\frac{x^{10-r}}{y^{10-r}}\right) \left(\frac{y^r}{x^r}\right) = \binom{10}{r} \frac{x^{10-r} y^r}{y^{10-r} x^r} = \binom{10}{r} \frac{x^{10-2r}}{y^{10-2r}} \]

### Step 3: Find the Middle Term
Since \(n = 10\) is even, the middle term will be the \(\frac{n}{2} + 1\)th term, which is the 6th term (\(T_6\)):
\[ T_6 = T_{5} \quad \text{(since we start counting from 0)} \]
Thus, we need to calculate \(T_5\) where \(r = 5\).

### Step 4: Substitute \(r = 5\) into the General Term
Now substituting \(r = 5\) into the general term:
\[ T_6 = \binom{10}{5} \frac{x^{10-2 \cdot 5}}{y^{10-2 \cdot 5}} = \binom{10}{5} \frac{x^{0}}{y^{0}} = \binom{10}{5} \]

### Step 5: Calculate \(\binom{10}{5}\)
Now we calculate \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252 \]

### Conclusion
Thus, the middle term in the expansion of \(\left(\frac{x}{y} + \frac{y}{x}\right)^{10}\) is:
\[ \boxed{252} \]