The term that is independent of x, in the expression `((3)/(2)x^(2)-(1)/(3x))^(9)` is

Options

A. `""^(9)C_(6)((3)/(2))^(5)(-(1)/(3))^(4)`
B. `""^(9)C_(3)((1)/(6))^(3)`
C. `""^(9)C_(4)((3)/(2))^(4)(-(1)/(3))^(5)`
D. `""^(9)C_(6)((3)/(2))^(6)(-(13)/(3))^(6)`

Answer

To find the term that is independent of \( x \) in the expression \( \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9 \), we can use the binomial theorem. The binomial theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, we can identify \( a = \frac{3}{2} x^2 \) and \( b = -\frac{1}{3x} \), and \( n = 9 \).

### Step 1: Write the general term
The general term \( T_r \) in the expansion is given by:

\[ T_r = \binom{9}{r} \left( \frac{3}{2} x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r \]

### Step 2: Simplify the general term
Now, let's simplify \( T_r \):

\[ T_r = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} x^{2(9-r)} \left( -\frac{1}{3} \right)^r x^{-r} \]

Combining the powers of \( x \):

\[ T_r = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18 - 3r} \]

### Step 3: Find the term independent of \( x \)
To find the term that is independent of \( x \), we need to set the exponent of \( x \) to zero:

\[ 18 - 3r = 0 \]

Solving for \( r \):

\[ 3r = 18 \implies r = 6 \]

### Step 4: Substitute \( r \) back into the general term
Now, we substitute \( r = 6 \) back into the general term \( T_r \):

\[ T_6 = \binom{9}{6} \left( \frac{3}{2} \right)^{9-6} \left( -\frac{1}{3} \right)^6 \]

### Step 5: Calculate the coefficients
Calculating each part:

1. \( \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \)
2. \( \left( \frac{3}{2} \right)^{3} = \frac{27}{8} \)
3. \( \left( -\frac{1}{3} \right)^{6} = \frac{1}{729} \)

### Step 6: Combine everything
Now, we combine these results:

\[ T_6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} \]

Calculating this gives:

\[ T_6 = 84 \cdot \frac{27}{8 \cdot 729} \]

### Step 7: Simplify the expression
Calculating \( 8 \cdot 729 = 5832 \):

\[ T_6 = \frac{84 \cdot 27}{5832} \]

Now, simplify \( \frac{84 \cdot 27}{5832} \):

\[ T_6 = \frac{2268}{5832} = \frac{1}{2.57} \approx 0.39 \text{ (exact value can be calculated)} \]

### Final Result
Thus, the term that is independent of \( x \) is:

\[ \frac{84 \cdot 27}{5832} = \frac{1}{2.57} \text{ (exact value can be calculated)} \]