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. Let's break down the solution step by step.

### Step 1: Write the General Term
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = \frac{3}{2} x^2 \), \( b = -\frac{1}{3x} \), and \( n = 9 \). Therefore, the general term can be expressed as:
\[ T_{r+1} = \binom{9}{r} \left( \frac{3}{2} x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r \]

### Step 2: Simplify the General Term
Now, we simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} x^{2(9-r)} \left( -\frac{1}{3} \right)^r x^{-r} \]
This simplifies to:
\[ T_{r+1} = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18 - 3r} \]

### Step 3: Find the Term Independent of \( x \)
To find the term that is independent of \( x \), we need the exponent of \( x \) to be 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:
\[ T_{7} = \binom{9}{6} \left( \frac{3}{2} \right)^{9-6} \left( -\frac{1}{3} \right)^6 \]
Calculating each part:
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
\[ \left( \frac{3}{2} \right)^{3} = \frac{27}{8} \]
\[ \left( -\frac{1}{3} \right)^6 = \frac{1}{729} \]

### Step 5: Combine All Parts
Putting it all together:
\[ T_{7} = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} \]
Calculating:
\[ T_{7} = 84 \cdot \frac{27}{8 \cdot 729} = \frac{84 \cdot 27}{5832} \]
Calculating \( 84 \cdot 27 = 2268 \):
\[ T_{7} = \frac{2268}{5832} = \frac{1}{2.5} = \frac{1}{2} \]

### Final Answer
Thus, the term that is independent of \( x \) in the expression is:
\[ \frac{1}{2} \]