To solve the problem [tex]\(\frac{x^{m+n} x y^{n-n}}{x^{m-n} x y^{n+n}}\)[/tex] given the values [tex]\(x=2\)[/tex], [tex]\(y=4\)[/tex], [tex]\(m=-1\)[/tex], and [tex]\(n=3\)[/tex], let's break down the expression step-by-step.
1. Substitute the values:
We are given:
- [tex]\(x = 2\)[/tex]
- [tex]\(y = 4\)[/tex]
- [tex]\(m = -1\)[/tex]
- [tex]\(n = 3\)[/tex]
The expression becomes:
[tex]\[
\frac{2^{(-1+3)} \cdot 2 \cdot 4^{(3-3)}}{2^{(-1-3)} \cdot 2 \cdot 4^{(3+3)}}
\][/tex]
2. Simplify the exponents within the numerator and denominator:
- [tex]\(m+n = -1+3 = 2\)[/tex]
- [tex]\(n-n = 3-3 = 0\)[/tex]
- [tex]\(m-n = -1-3 = -4\)[/tex]
- [tex]\(n+n = 3+3 = 6\)[/tex]
Substituting these, the expression becomes:
[tex]\[
\frac{2^{2} \cdot 2 \cdot 4^{0}}{2^{-4} \cdot 2 \cdot 4^{6}}
\][/tex]
3. Evaluate the exponents:
- [tex]\(2^2 = 4\)[/tex]
- [tex]\(4^0 = 1\)[/tex] (since any number raised to the power of 0 is 1)
- [tex]\(2^{-4} = \frac{1}{2^4} = \frac{1}{16}\)[/tex]
- [tex]\(4^6 = 4096\)[/tex] (since [tex]\(4^6 = (2^2)^6 = 2^{12} = 4096\)[/tex])
Now, substitute these back into the expression:
[tex]\[
\frac{4 \cdot 2 \cdot 1}{\frac{1}{16} \cdot 2 \cdot 4096}
\][/tex]
4. Simplify the terms:
Numerator:
[tex]\[
4 \cdot 2 \cdot 1 = 8
\][/tex]
Denominator:
[tex]\[
\frac{1}{16} \cdot 2 \cdot 4096 = \frac{4096 \cdot 2}{16} = \frac{8192}{16} = 512
\][/tex]
Therefore, the expression simplifies to:
[tex]\[
\frac{8}{512}
\][/tex]
5. Calculate the final value:
Simplify [tex]\(\frac{8}{512}\)[/tex]:
[tex]\[
\frac{8}{512} = \frac{1}{64} = 0.015625 = 3.814697265625 \times 10^{-6}
\][/tex]
Hence, the value of the given expression is:
[tex]\[
3.814697265625 \times 10^{-6}
\][/tex]
So, [tex]\((\text{Numerator: } 8, \text{ Denominator: } 2097152.0, \text{ Value: } 3.814697265625 \times 10^{-6})\)[/tex].
