Question

Question 26

Fill in the blanks to make the statement true
${\left(\frac{-1}{4}\right)}^3\times \text{\_\_}={\left(\frac{-1}{4}\right)}^{11}$

Answer 1

Let ${\left(\frac{-1}{4}\right)}^3\times x={\left(\frac{-1}{4}\right)}^{11}\Rightarrow x=\frac{{\left(\frac{-1}{4}\right)}^{11}}{{\left(\frac{-1}{4}\right)}^3}\Rightarrow x={\left(\frac{-1}{4}\right)}^{11-3}[\because a^m÷a^n=a^{m-n},m>n]\Rightarrow x={\left(\frac{-1}{4}\right)}^8$.

$\therefore {(-\frac{1}{4})}^3\times {(-\frac{1}{4})}^8={(-\frac{1}{4})}^{11}$.