Question

1. \\(\frac{(2n^{-3})^{2}}{2m^{0}n^{4} cdot nm^{3}}\\)

Explanation:

Step1: Simplify the numerator using power of a power rule

The power of a power rule states that \((a^m)^n = a^{mn}\). For the numerator \((2n^{-3})^2\), we apply this rule:
\((2n^{-3})^2 = 2^2 \cdot (n^{-3})^2 = 4n^{-6}\)

Step2: Simplify the denominator using product of powers rule and zero exponent rule

First, recall that \(m^0 = 1\) (zero exponent rule). Then, for the denominator \(2m^0n^4 \cdot nm^3\), we use the product of powers rule \(a^m \cdot a^n = a^{m + n}\) for the same base.
First, simplify \(m\) terms: \(m^0 \cdot m^3 = m^{0 + 3}=m^3\)
Then, simplify \(n\) terms: \(n^4 \cdot n = n^{4+ 1}=n^5\)
And the coefficient is \(2\). So the denominator becomes \(2\times1\times n^5\times m^3=2m^3n^5\)

Step3: Divide the numerator by the denominator

Now we have \(\frac{4n^{-6}}{2m^3n^5}\). First, simplify the coefficients: \(\frac{4}{2}=2\)
Then, use the quotient of powers rule \(a^m\div a^n=a^{m - n}\) for the \(n\) terms: \(n^{-6}\div n^5=n^{-6-5}=n^{-11}\)
The \(m\) term remains as \(m^3\) in the denominator. So we have \(2\times\frac{n^{-11}}{m^3}\)
Recall that \(a^{-n}=\frac{1}{a^n}\), so \(n^{-11}=\frac{1}{n^{11}}\). Thus, the expression becomes \(\frac{2}{m^3n^{11}}\)

Answer:

\(\boxed{\dfrac{2}{m^3n^{11}}}\)