Examine whether or not there is any term containing `x^(9)` in the expansion of `(2x^(2) - (1)/(x))^(20)`

Answer

To determine whether there is a term containing \( x^9 \) in the expansion of \( (2x^2 - \frac{1}{x})^{20} \), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2x^2 \), \( b = -\frac{1}{x} \), and \( n = 20 \).

2. **Write the General Term**:
For our specific case:
\[ T_{r+1} = \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]
Simplifying this, we get:
\[ T_{r+1} = \binom{20}{r} (2^{20-r} (x^2)^{20-r}) \left(-1\right)^r \left(\frac{1}{x^r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{20}{r} (-1)^r 2^{20-r} x^{2(20-r) - r} \]

3. **Simplify the Exponent of \( x \)**:
The exponent of \( x \) in the term is:
\[ 2(20 - r) - r = 40 - 2r - r = 40 - 3r \]
Therefore, we have:
\[ T_{r+1} = \binom{20}{r} (-1)^r 2^{20-r} x^{40 - 3r} \]

4. **Set the Exponent Equal to 9**:
We want to find \( r \) such that:
\[ 40 - 3r = 9 \]
Rearranging gives:
\[ 3r = 40 - 9 = 31 \]
Thus:
\[ r = \frac{31}{3} \]

5. **Check if \( r \) is an Integer**:
Since \( r = \frac{31}{3} \) is not an integer, there is no valid \( r \) that satisfies this equation.

6. **Conclusion**:
Since there is no integer value of \( r \) that results in a term containing \( x^9 \), we conclude that there is no term containing \( x^9 \) in the expansion of \( (2x^2 - \frac{1}{x})^{20} \).

### Final Answer:
No such term exists.