Question

Simplify ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}÷\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+{\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)}^0$

Answer 1

Simplify the given expression:

$\begin{array}{l}{\left(\frac{64}{125}\right)}^{-\frac{2}{3}}÷\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+{\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)}^0\\ ={\left(\frac{4^3}{5^3}\right)}^{-\frac{2}{3}}÷\frac{1}{{\left(\frac{4^4}{5^4}\right)}^{\frac{1}{4}}}+1\left(\because a^0=1\right)\\ ={\left(\frac{4}{5}\right)}^{-2}÷\frac{1}{\left(\frac{4}{5}\right)}+1\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ ={\left(\frac{4}{5}\right)}^{-2}\times {\left(\frac{4}{5}\right)}^1+1\\ ={\left(\frac{4}{5}\right)}^{-2+1}+1\\ ={\left(\frac{4}{5}\right)}^{-1}+1\\ =\frac{5}{4}+1\left(\because a^{-m}=\frac{1}{a^m}\right)\\ =\frac{5+4}{4}\\ =\frac{9}{4}\\ =2\frac{1}{4}\end{array}$

Hence,the value of the given expression is ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}÷\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+{\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)}^0=2\frac{1}{4}$