Which term of the expansion `(sqrt((x)/(3))-(sqrt(3))/(2x))^(12)`is independent of x.

Answer

To find the term in the expansion of \((\sqrt{\frac{x}{3}} - \frac{\sqrt{3}}{2x})^{12}\) that is independent of \(x\), we can follow these steps:

### Step 1: Identify \(a\), \(b\), and \(n\)
We can rewrite the expression as:
\[ a = \sqrt{\frac{x}{3}} = \frac{\sqrt{x}}{\sqrt{3}}, \quad b = -\frac{\sqrt{3}}{2x}, \quad n = 12 \]

### Step 2: Write the general term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Substituting the values of \(a\), \(b\), and \(n\):
\[ T_{r+1} = \binom{12}{r} \left(\frac{\sqrt{x}}{\sqrt{3}}\right)^{12-r} \left(-\frac{\sqrt{3}}{2x}\right)^r \]

### Step 3: Simplify the general term
Now, we simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{12}{r} \left(\frac{x^{1/2}}{3^{1/2}}\right)^{12-r} \left(-\frac{\sqrt{3}}{2} \cdot x^{-1}\right)^r \]
This can be rewritten as:
\[ T_{r+1} = \binom{12}{r} \cdot \frac{(-1)^r \cdot 3^{r/2}}{2^r} \cdot \frac{x^{(12-r)/2}}{3^{(12-r)/2}} \cdot x^{-r} \]
Combining the \(x\) terms:
\[ T_{r+1} = \binom{12}{r} \cdot \frac{(-1)^r \cdot 3^{r/2}}{2^r \cdot 3^{(12-r)/2}} \cdot x^{\frac{12-r}{2} - r} \]
This simplifies to:
\[ T_{r+1} = \binom{12}{r} \cdot \frac{(-1)^r}{2^r} \cdot \frac{3^{(r - (12-r))/2}}{1} \cdot x^{\frac{12 - 3r}{2}} \]

### Step 4: Find the term independent of \(x\)
For the term to be independent of \(x\), the exponent of \(x\) must equal zero:
\[ \frac{12 - 3r}{2} = 0 \]
Solving for \(r\):
\[ 12 - 3r = 0 \implies 3r = 12 \implies r = 4 \]

### Step 5: Find the specific term
Now, we substitute \(r = 4\) back into the general term to find the specific term:
\[ T_{4+1} = T_5 = \binom{12}{4} \cdot \frac{(-1)^4 \cdot 3^{(4 - 8)/2}}{2^4} \]
Calculating \(\binom{12}{4}\):
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]
Thus, the term becomes:
\[ T_5 = 495 \cdot \frac{1 \cdot 3^{-2}}{16} = 495 \cdot \frac{1}{9 \cdot 16} = \frac{495}{144} \]

### Conclusion
The fifth term of the expansion is independent of \(x\).