The value of `(4^nxx20^(m-1)xx12^(m-n)xx15^(m+n-2))/(16^mxx5^(2m+n)xx9^(m-1))` s-

Options

A. `1/50`
B. `1/500`
C. `1/100`
D. `1/5`

Answer

To solve the expression

\[ \frac{4^n \times 20^{m-1} \times 12^{m-n} \times 15^{m+n-2}}{16^m \times 5^{2m+n} \times 9^{m-1}}, \]

we will break down each term into its prime factors and simplify the expression step by step.

### Step 1: Factor each term into prime factors

1. **For \(4^n\)**:
\[ 4 = 2^2 \implies 4^n = (2^2)^n = 2^{2n} \]

2. **For \(20^{m-1}\)**:
\[ 20 = 2^2 \times 5 \implies 20^{m-1} = (2^2 \times 5)^{m-1} = 2^{2(m-1)} \times 5^{m-1} \]

3. **For \(12^{m-n}\)**:
\[ 12 = 2^2 \times 3 \implies 12^{m-n} = (2^2 \times 3)^{m-n} = 2^{2(m-n)} \times 3^{m-n} \]

4. **For \(15^{m+n-2}\)**:
\[ 15 = 3 \times 5 \implies 15^{m+n-2} = (3 \times 5)^{m+n-2} = 3^{m+n-2} \times 5^{m+n-2} \]

5. **For \(16^m\)**:
\[ 16 = 2^4 \implies 16^m = (2^4)^m = 2^{4m} \]

6. **For \(5^{2m+n}\)**: This remains as is.

7. **For \(9^{m-1}\)**:
\[ 9 = 3^2 \implies 9^{m-1} = (3^2)^{m-1} = 3^{2(m-1)} = 3^{2m-2} \]

### Step 2: Substitute back into the expression

Now substituting these factorizations back into the original expression, we have:

\[ \frac{2^{2n} \times 2^{2(m-1)} \times 2^{2(m-n)} \times 3^{m-n} \times 5^{m-1} \times 3^{m+n-2} \times 5^{m+n-2}}{2^{4m} \times 5^{2m+n} \times 3^{2m-2}}. \]

### Step 3: Combine the powers

Now we combine the powers of each base:

1. **For base \(2\)**:
\[ \text{Numerator: } 2^{2n + 2(m-1) + 2(m-n)} = 2^{2n + 2m - 2 + 2m - 2n} = 2^{4m - 2} \]
\[ \text{Denominator: } 2^{4m} \]
\[ \text{Power of } 2: \quad 2^{(4m - 2) - 4m} = 2^{-2} \]

2. **For base \(3\)**:
\[ \text{Numerator: } 3^{m-n + m+n-2} = 3^{2m - 2} \]
\[ \text{Denominator: } 3^{2m-2} \]
\[ \text{Power of } 3: \quad 3^{(2m - 2) - (2m - 2)} = 3^0 = 1 \]

3. **For base \(5\)**:
\[ \text{Numerator: } 5^{m-1 + m+n-2} = 5^{2m + n - 3} \]
\[ \text{Denominator: } 5^{2m+n} \]
\[ \text{Power of } 5: \quad 5^{(2m + n - 3) - (2m + n)} = 5^{-3} \]

### Step 4: Combine the results

Now we combine the results:

\[ \frac{2^{-2} \times 1 \times 5^{-3}}{1} = \frac{1}{2^2 \times 5^3} = \frac{1}{4 \times 125} = \frac{1}{500}. \]

### Final Answer

Thus, the value of the expression is

\[ \frac{1}{500}. \]