Find the constant term in the expansion of `(sqrtx+1/(3x^2))^10`.

Explanation

To find the constant term in the expansion of \((\sqrt{x} + \frac{1}{3x^2})^{10}\), we will use the Binomial Theorem. Let's break it down step by step.

### Step 1: Identify the general term
The general term in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
where \(n\) is the exponent, \(a\) is the first term, and \(b\) is the second term.

In our case:
- \(n = 10\)
- \(a = \sqrt{x}\)
- \(b = \frac{1}{3x^2}\)

Thus, the general term \(T_{r+1}\) becomes:
\[ T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{1}{3x^2}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \cdot \frac{1}{3^r} \cdot \frac{1}{(x^2)^r} \]
This can be rewritten as:
\[ T_{r+1} = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10-r}{2}} \cdot x^{-2r} \]
Combining the powers of \(x\):
\[ T_{r+1} = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10-r}{2} - 2r} \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10 - r - 4r}{2}} = \binom{10}{r} \cdot \frac{1}{3^r} \cdot x^{\frac{10 - 5r}{2}} \]

### Step 3: Set the exponent of \(x\) to zero
To find the constant term, we need the exponent of \(x\) to be zero:
\[ \frac{10 - 5r}{2} = 0 \]
Multiplying through 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 to find the constant 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, we have:
\[ T_{3} = 45 \cdot \frac{1}{9} = 5 \]

### Conclusion
The constant term in the expansion of \((\sqrt{x} + \frac{1}{3x^2})^{10}\) is **5**.