Find the term independent of `x` in the expansion of the following binomials:
(ii) `( sqrt((x)/(3) ) - sqrt(3)/(2x ))^(12)`

Answer

To find the term independent of \( x \) in the expansion of \( \left( \sqrt{\frac{x}{3}} - \frac{\sqrt{3}}{2x} \right)^{12} \), we can 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, \( a = \sqrt{\frac{x}{3}} \) and \( b = -\frac{\sqrt{3}}{2x} \), and \( n = 12 \).

### Step 2: Write the General Term
Substituting the values into the formula, we get:

\[ T_{r+1} = \binom{12}{r} \left( \sqrt{\frac{x}{3}} \right)^{12-r} \left( -\frac{\sqrt{3}}{2x} \right)^r \]

### Step 3: Simplify the General Term
Now, we simplify the general term:

\[ T_{r+1} = \binom{12}{r} \left( \frac{x^{1/2}}{3^{1/2}} \right)^{12-r} \left( -\frac{\sqrt{3}}{2} \right)^r \cdot x^{-r} \]

This can be rewritten as:

\[ T_{r+1} = \binom{12}{r} (-1)^r \frac{3^{r/2}}{2^r} \cdot \frac{x^{(12-r)/2}}{3^{(12-r)/2}} \cdot x^{-r} \]

Combining the powers of \( x \):

\[ T_{r+1} = \binom{12}{r} (-1)^r \frac{3^{r/2}}{2^r} \cdot x^{\frac{12-r}{2} - r} \]

### Step 4: Find the Power of \( x \)
The exponent of \( x \) in \( T_{r+1} \) is:

\[ \frac{12 - r}{2} - r = \frac{12 - r - 2r}{2} = \frac{12 - 3r}{2} \]

### Step 5: Set the Exponent to Zero
To find the term independent of \( x \), we set the exponent equal to zero:

\[ \frac{12 - 3r}{2} = 0 \]

Multiplying through by 2:

\[ 12 - 3r = 0 \implies 3r = 12 \implies r = 4 \]

### Step 6: Substitute \( r \) Back into the General Term
Now, substitute \( r = 4 \) back into the general term to find the independent term:

\[ T_{5} = \binom{12}{4} (-1)^4 \frac{3^{4/2}}{2^4} = \binom{12}{4} \frac{3^2}{16} \]

Calculating \( \binom{12}{4} \):

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]

### Step 7: Calculate the Independent Term
Now, substituting back:

\[ T_{5} = 495 \cdot \frac{9}{16} = \frac{4455}{16} \]

### Final Answer
The term independent of \( x \) in the expansion is:

\[ \frac{4455}{16} \]