Find the coefficient of
`x^(-7)" in "((2x^(2))/(3)-(5)/(4x^(5)))^(7)`

Answer

To find the coefficient of \( x^{-7} \) in the expression

\[ \left( \frac{2x^2}{3} - \frac{5}{4x^5} \right)^7, \]

we will use the Binomial Theorem, which states that:

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

### Step 1: Identify \( a \) and \( b \)

In our case, we can identify:

- \( a = \frac{2x^2}{3} \)
- \( b = -\frac{5}{4x^5} \)

### Step 2: Write the general term

The general term \( T_{r+1} \) in the expansion is given by:

\[ T_{r+1} = \binom{7}{r} \left( \frac{2x^2}{3} \right)^{7-r} \left( -\frac{5}{4x^5} \right)^r. \]

### Step 3: Simplify the general term

Now, we simplify \( T_{r+1} \):

\[ T_{r+1} = \binom{7}{r} \left( \frac{2^{7-r} x^{2(7-r)}}{3^{7-r}} \right) \left( -\frac{5^r}{4^r x^{5r}} \right). \]

Combining these, we have:

\[ T_{r+1} = \binom{7}{r} \cdot \frac{2^{7-r} \cdot (-5)^r}{3^{7-r} \cdot 4^r} \cdot x^{2(7-r) - 5r}. \]

### Step 4: Find the exponent of \( x \)

The exponent of \( x \) in \( T_{r+1} \) is given by:

\[ 2(7-r) - 5r = 14 - 2r - 5r = 14 - 7r. \]

### Step 5: Set the exponent equal to -7

We need to find \( r \) such that:

\[ 14 - 7r = -7. \]

Solving for \( r \):

\[ 14 + 7 = 7r \implies 21 = 7r \implies r = 3. \]

### Step 6: Substitute \( r \) back into the general term

Now, we substitute \( r = 3 \) into the general term:

\[ T_{4} = \binom{7}{3} \cdot \frac{2^{7-3} \cdot (-5)^3}{3^{7-3} \cdot 4^3}. \]

### Step 7: Calculate \( T_{4} \)

Calculating each component:

1. \( \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)
2. \( 2^{4} = 16 \)
3. \( (-5)^{3} = -125 \)
4. \( 3^{4} = 81 \)
5. \( 4^{3} = 64 \)

Putting it all together:

\[ T_{4} = 35 \cdot \frac{16 \cdot (-125)}{81 \cdot 64}. \]

### Step 8: Simplify the coefficient

Calculating the numerator:

\[ 16 \cdot (-125) = -2000. \]

Now, substituting back:

\[ T_{4} = 35 \cdot \frac{-2000}{5184}. \]

Calculating \( 35 \cdot -2000 = -70000 \), and simplifying:

\[ T_{4} = \frac{-70000}{5184}. \]

### Final Coefficient

Thus, the coefficient of \( x^{-7} \) in the expression is:

\[ \frac{-70000}{5184}. \]