Question

(28d³k² + d²k² - 4dk²)(4dk²)⁻¹

○ 96d² + 4d - 16
○ 7d⁵k⁴ + (d³k⁴)/4 - d²k⁴
○ 7d² + d/4 - 1
○ 1/(7d²) + d/4 - 1

Explanation:

Step1: Rewrite the inverse

Recall that \( (ab)^{-1}=\frac{1}{ab} \), so \( (4dk^2)^{-1}=\frac{1}{4dk^2} \). The expression becomes \( (28d^3k^2 + d^2k^2 - 4dk^2)\times\frac{1}{4dk^2} \).

Step2: Distribute the fraction

Distribute \( \frac{1}{4dk^2} \) to each term in the parentheses:

• For the first term: \( 28d^3k^2\times\frac{1}{4dk^2}=\frac{28d^3k^2}{4dk^2} \)
• For the second term: \( d^2k^2\times\frac{1}{4dk^2}=\frac{d^2k^2}{4dk^2} \)
• For the third term: \( - 4dk^2\times\frac{1}{4dk^2}=\frac{-4dk^2}{4dk^2} \)

Step3: Simplify each term

• Simplify the first term: \( \frac{28d^3k^2}{4dk^2}=\frac{28}{4}d^{3 - 1}k^{2 - 2}=7d^2 \) (using \( \frac{a^m}{a^n}=a^{m - n} \) and \( k^0 = 1,k^2/k^2 = 1 \))
• Simplify the second term: \( \frac{d^2k^2}{4dk^2}=\frac{1}{4}d^{2 - 1}k^{2 - 2}=\frac{d}{4} \)
• Simplify the third term: \( \frac{-4dk^2}{4dk^2}=- 1 \)

Step4: Combine the simplified terms

Combining the three simplified terms, we get \( 7d^2+\frac{d}{4}-1 \).

Answer:

\( 7d^2+\frac{d}{4}-1 \) (the third option)