The term independent of x in the expansion of `[sqrt(((x)/(3)))+sqrt(((3)/(2x^(2))))]^(10)` is

Options

A. none
B. `""^(10)C_(1)`
C. `5//12`
D. 1

Answer

To find the term independent of \( x \) in the expansion of

\[ \left( \sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} \right)^{10} \]

we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

In our case, \( a = \sqrt{\frac{x}{3}} \) and \( b = \sqrt{\frac{3}{2x^2}} \), and \( n = 10 \).

### Step 2: Write the general term for our specific case
Substituting the values of \( a \), \( b \), and \( n \):

\[ T_{r+1} = \binom{10}{r} \left( \sqrt{\frac{x}{3}} \right)^{10-r} \left( \sqrt{\frac{3}{2x^2}} \right)^{r} \]

### Step 3: Simplify the expression
Now, simplify \( T_{r+1} \):

\[ T_{r+1} = \binom{10}{r} \left( \frac{x^{1/2}}{3^{1/2}} \right)^{10-r} \left( \frac{\sqrt{3}}{\sqrt{2} \cdot x} \right)^{r} \]

This can be rewritten as:

\[ T_{r+1} = \binom{10}{r} \frac{x^{(10-r)/2}}{3^{5/2}} \cdot \frac{3^{1/2}}{(2^{1/2} \cdot x^{r})} \]

Combining the powers of \( x \):

\[ T_{r+1} = \binom{10}{r} \cdot \frac{3^{(1/2)(1-r) - (1/2)(r)}}{2^{1/2}} \cdot x^{(10-r)/2 - r} \]

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

\[ \frac{10 - r}{2} - r = \frac{10 - r - 2r}{2} = \frac{10 - 3r}{2} \]

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

\[ \frac{10 - 3r}{2} = 0 \]

Multiplying through by 2:

\[ 10 - 3r = 0 \]

Solving for \( r \):

\[ 3r = 10 \implies r = \frac{10}{3} \]

### Step 6: Check if \( r \) is an integer
Since \( r = \frac{10}{3} \) is not an integer, there is no term in the expansion that is independent of \( x \).

### Conclusion
Thus, the term independent of \( x \) in the expansion does not exist.