Find the terms independent of x in the expansion of `(sqrt(x)+(1)/(3x^(2)))^(10)`

Explanation

To find the term independent of \( x \) in the expansion of \( \left( \sqrt{x} + \frac{1}{3x^2} \right)^{10} \), 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, \( a = \sqrt{x} \), \( b = \frac{1}{3x^2} \), and \( n = 10 \).

### Step 2: Write the General Term
Substituting the values into the general term formula, we have:
\[ T_{r+1} = \binom{10}{r} \left( \sqrt{x} \right)^{10-r} \left( \frac{1}{3x^2} \right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} \left( \sqrt{x} \right)^{10-r} \cdot \frac{1}{3^r} \cdot \frac{1}{x^{2r}} \]
\[ = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10-r}{2}} \cdot x^{-2r} \]
\[ = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10-r - 4r}{2}} = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10 - 5r}{2}} \]

### Step 3: Set the Exponent of \( x \) to Zero
To find the term that is independent of \( x \), we need the exponent of \( x \) to be zero:
\[ \frac{10 - 5r}{2} = 0 \]
Multiplying both sides by 2 gives:
\[ 10 - 5r = 0 \]
Solving for \( r \):
\[ 5r = 10 \implies r = 2 \]

### Step 4: Substitute \( r \) Back into the General Term
Now, we substitute \( r = 2 \) back into the general term:
\[ T_{3} = \binom{10}{2} \cdot \frac{1}{3^2} \cdot x^{0} \]
Calculating \( \binom{10}{2} \):
\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus:
\[ T_{3} = 45 \cdot \frac{1}{9} = 5 \]

### Conclusion
The term independent of \( x \) in the expansion is:
\[ \boxed{5} \]