Question

$\left(\dfrac{-4m^{7}k^{8}}{h^{2}m^{5}k^{7}}\
ight)^{2}$

Explanation:

Step1: Simplify the fraction inside the parentheses

First, simplify the coefficients and the variables with the same base separately. For the coefficient: $\frac{-4}{1}=-4$. For the variable \(m\): use the rule \( \frac{m^a}{m^b}=m^{a - b} \), so \( \frac{m^7}{m^5}=m^{7 - 5}=m^2 \). For the variable \(k\): \( \frac{k^8}{k^7}=k^{8 - 7}=k^1 = k \). For the variable \(h\): it remains as \(h^{-2}\) (since it's in the denominator, we can write it as \(h^{-2}\) using the rule \( \frac{1}{h^n}=h^{-n} \)). So the fraction inside the parentheses becomes \( \frac{-4m^2k}{h^2} \) (or \( -4m^2k h^{-2} \)).

Step2: Apply the power of 2 to the simplified fraction

Now, raise each part to the power of 2. For the coefficient: \( (-4)^2 = 16 \). For the variable \(m\): \( (m^2)^2 = m^{2\times2}=m^4 \) (using the rule \( (a^m)^n=a^{m\times n} \)). For the variable \(k\): \( (k)^2 = k^2 \). For the variable \(h\): \( (h^{-2})^2 = h^{-2\times2}=h^{-4}=\frac{1}{h^4} \) (or we can keep it with positive exponents in the denominator). Putting it all together, we have \( \frac{16m^4k^2}{h^4} \).

Answer:

\(\dfrac{16m^{4}k^{2}}{h^{4}}\)